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A Little eBookof
Calculus Tests
Interactive Self -Tests and Practice Exams
Note: Text displayed in this reddish colour is a link.
Last update on October 15, 2010
Contents
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Contents
1 Limits 3
Self-Test 1.1 • Self-Test 1.2
2 Differentiation 5
Using the Definition of the Derivative 2.1 • Polynomials 2.2 •Product & Quotient Rules 2.3 • Chain/Power Rule 2.4 •Implicit Function Differentiation 2.5
3 Quiz I Practice (Limits and Using the Definition of the Derivative) 10
Quiz 3.1 • Quiz 3.2 • Quiz 3.3 • Quiz 3.4
4 Quiz I Practice (Limits, Definition of the Derivative, Product and Quotient Rules) 14
Quiz 4.1
5 Applications of Differentiation 15
Tangents & Normals 5.1 • Related Rates 5.2
6 Midterm Practice (Differentiation and Applications) 17
Quiz 6.1 • Quiz 6.2
7 Integration 19
Indefinite Integration 7.1 • Definite Integration 7.2
8 Quiz 3 Practice (Integration) 21
Quiz 8.1 • Quiz 8.2 • Quiz 8.3
9 Applications of Integration 24
Area 9.1 • Work 9.2
Solutions to Quizzes 26
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1. Limits
1.1. Self-Test
To start, press
Evaluate the limits shown below. (If the limit does not exist, enter dne or does not exist.)
1. limx→42
x21
=
2. limx→−2
3− 4x =
3. limx→3
3x√
x2 − 9 =
4. limx→2
x2 =
5. limx→−2
x2 − 4x + 2
=
6. limx→2
x2 − 4x + 2
=
7. limx→2
∣∣x− 2∣∣
x− 2=
8. limx→9
√x− 3
9− x=
9. limx→2
(x− 2)√
x2 − 11 =
10. limx→2
x3 − 8x2 − 4
=
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1.2. Self-Test
To start, press
Evaluate the limits shown below. (If the limit does not exist, enter dne or does not exist.)
1. limx→4
√x2 + 9x + 3
=
2. limx→2
(x− 3)3 +4
2x2 =
3. limt→−1
3t2 − 4t− 1
=
4. limx→2
x2 + x− 6x2 − x− 2
=
5. limx→∞
1x2 =
6. limx→2
√4− x2 =
7. limx→2
1x −
12
x− 2=
8. limx→2
x3 − 8x− 2
=
9. limx→−1
x3 + 1x + 1
=
10. limx→2
x2 + 3t− 10x2 − 4
=
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2. Differentiation
2.1. Using the Definition of the Derivative
To start, press
Use the definition to find the derivative of the following functions, or to evaluate for the given variablevalue as indicated by the question.
Enter your input as you would to a calculator or spreadsheet, e.g. enter3
4√
xas 3/(4x^0.5).
1. f (x) = 3x− x2. f ′(x) =
2. f (t) = t3 + t2 + t. f ′(t) =
3. y = 2x +2x
. f ′(x) =
4. f (s) =s
2s + 1. f ′(s) =
5. y =√
x + 1. y′(x) =
6. s(t) = 42t− 5t2. Find the value of t
where the tangent to s(t) is horizontal. t =
7. f (x) =√
4− x2.
Find x such that the slope of f (x) is −1. x =
8. a)d
dx
(1√x
)=
b) The tangent line at
(4,
12
)is y =
9. f (x) =1− x1 + x
. f ′(x) =
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2.2. Polynomials
To start, press
Enter your input as you would to a calculator or spreadsheet, e.g. enter3
4√
xas 3/(4x^0.5).
1. f (x) = 4x3 − 3x2 + 2x− 1. f ′(x) =
2. s(t) = (3t)4 − (2t)5. s′(t) =
3. f (x) =√
2x. f ′(x) =
4. g(t) = π(√
4t2 − 3√
8t3)(√
4t2 +3√
8t3). g′(t) =
5. f (x) =(1− x3)(x3 + 1
). f ′(x) =
6. f (x) = x2√x− 1√x+ x3√
x3. f ′(x) =
7. A ball is thrown vertically upwards. The ball’s elevation in metresafter t seconds is given by h(t) = 22t− 5t2. The velocity(instantaneous change of elevation) of the ball is given by h′(t).Determine values (without units) for:
a) the ball’s velocity, in m/s, after 1.5 seconds =
b) the time, in seconds, to reach the highest point =
b) the maximum height, in metres, of the ball =
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2.3. Product & Quotient Rules
To start, press Input style:3y2
4√
xis entered as 3/(4x^0.5), etc.
Use the product rule to find the following derivatives:
1. h(t) = (2t + 3)(4t− 5). h′(t) =
2. y(x) =(2x3 − 3
)(3x2 + 2
). y′(x) =
3. g(x) =(3x2 + 2x + 1
)2. g′(x) =
4. f (s) =(s2 + s + 1
)(s2 − s + 1
). f ′(s) =
5. y =
(2√
x +3x
)(3√
x− 2x
). y′ =
Use the quotient rule to find the following derivatives:
6. y(x) =x− 3x + 3
. y′(x) =
7. g(t) =t2
t + 1. g′(t) =
8. y(x) =x2 + 1
2x2 + 1. y′(x) =
9. p(s) =s2 + 1√
s− 1. p′(s) =
10. y(x) =1
1 + 1x
. y′ =
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2.4. Chain/Power Rule
To start, press . Input style:3y2
4√
xis entered as 3y^2/(4x^0.5).
1. f (x) = (3x + 2)3. Then f ′(x) =
2. g(t) = (10− 3t2)5. Then g′(t) =
3. h(x) =1
(x2 + 2)2 . Then h′(x) =
4. y =√
3x2 − 2x + 1. Thendydx
=
5. y =(
1 + x3)1/3
. Thendydx
=
6. s(t) =
√5t
. Then s′(t) =
7. y = x3(
2x2 − 3)4
. Then y′ =
8. y = (2x + 1)3(3x + 1)2. Then y′ =
9. s(t) =(
1 + t2/3)3/2
. Thendsdt
=
10. g(y) =[1 + (2y + 3)4
]5. Then g′(y) =
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2.5. Implicit Function Differentiation
To start, press . Input style:3y2
4√
xis entered as 3y^2/(4x^0.5).
Multiple answers to a question part should be separated by commas, e.g.
1. x + xy + y = 0.
a) Use implicit differentiation to find y′.
b) Find y’ at (−2,−2), (0, 0) and at (2,−2/3).
c) Rearrange x + xy + y = 0 so that y = g(x) is an explicit function
of x. That is, isolate the variable y. y = g(x) =
d) Differentiate g(x), the result from c) above.dgdx
=
e) Find g’ at (−2,−2), (0, 0) and at (2,−2/3).
2. x2y2 − x3 + y3 = 0.dydx
=
3. x2y2 = x3 + y3.dydx
=
4. (x− 2)2 + (y− 1)2 = 25 is the equation of the circle with centre (2, 1) and radius 5.
a) The derivativedydx
=
5. (x + y)2 + (x− y)2 = x3 + y3.dydx
=
6. x2y− xy2 + x2 + y2 = 0. y′ =
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3. Quiz I Practice (Limits and Using the Definition of theDerivative)
3.1. Quiz
To start, press
1. Evaluate the limit or, if it does not exist, write dne: limx→2
2x2 − x + 3 =
2. Evaluate the limit or, if it does not exist, write dne: limt→42
√t + 7 =
3. Evaluate the limit or, if it does not exist, write dne: limx→−1
2− xx2 + 1
=
4. Evaluate the limit or, if it does not exist, write dne: limh→√
3
√h2 − 3 =
5. Evaluate the limit or, if it does not exist, write dne: limx→−2/3
6 + 9x|3x + 2| =
6. Evaluate the limit or, if it does not exist, write dne: lims→−1
1 + 3s + s2
s=
7. Evaluate the limit or, if it does not exist, write dne: limt→5
t2 − 4t− 5t− 5
=
In the following two questions, use the definition of the derivative to find the derivative of the function.
8. The derivative of s(t) = 3t− 5t2 is given by s′(t) =
9. The derivative of f (x) =√
x− 2 is given by f ′(x) =
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3.2. Quiz
To start, press
1. Evaluate the limit or, if it does not exist, write dne: limx→2
2(x− 1)2 − 2(x + 1) =
2. Evaluate the limit or, if it does not exist, write dne: limx→1
√x2 + 4x + 2
=
3. Evaluate the limit or, if it does not exist, write dne: lims→2
4− s2
2− s=
4. Evaluate the limit or, if it does not exist, write dne: limx→2
x2 + 5x− 14x2 + x− 6
=
5. Evaluate the limit or, if it does not exist, write dne: limx→1
∣∣x2 − 1∣∣
x− 1=
6. Evaluate the limit or, if it does not exist, write dne: limt→2
√(t− 2)3
t− 2=
In the following two questions, use the definition of the derivative to find the derivative of the function.
7. The derivative of y(x) =1
1 + xis given by y′(x) =
8. a)d
dx(2x + 3)2 =
b) The equation of the tangent line with slope 4 is y =
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3.3. Quiz
To start, press
Evaluate the limits shown below. (If the limit does not exist, enter dne or does not exist.)
1. limx→−1
(3x + 2)3 − x2 =
2. limh→2
√h2 − 4 =
3. lims→1/3
√1 + 9s2
1 + 3s=
4. limp→0
3p2 − 2pp
=
5. limy→−2
8 + y3
y + 2=
6. limn→2
n2 − 3n + 2n2 + n− 6
=
Use the definition of a derivative to find the following derivatives.
7. f (x) = 3x2 + 2x. f ′(x) =
8. s(t) =√
t + 7. s′(t) =
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3.4. Quiz
To start, press
Evaluate the limits shown below. (If the limit does not exist, enter dne or does not exist.)
1. limx→−1
(2x + 3)2 − x3 =
2. limt→3
√t2 + 9 =
3. lims→2
√4− s2
2− s=
4. limq→0
q + 5q2
5q=
5. limw→1
w3 − 1w− 1
=
6. limm→1
m2 − 3m− 4m2 − 3m + 2
=
Use the definition of a derivative to find the following derivatives.
7. s(t) = 2t2 + 3t. s′(t) =
8. f (x) =√
x− 5. f ′(x) =
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4. Quiz I Practice (Limits, Definition of the Derivative, Productand Quotient Rules)
4.1. Quiz
To start, press
1. Evaluate the limit or, if it does not exist, write dne: limx→2
x2 + 3x− 4 =
2. Evaluate the limit or, if it does not exist, write dne: limh→−4
√h2 + 9 =
3. Evaluate the limit or, if it does not exist, write dne: limx→2
√x2 − 4 =
4. Evaluate the limit or, if it does not exist, write dne: limp→3
p2 − 9p− 3
=
5. Evaluate the limit or, if it does not exist, write dne: limy→−4
y2 + y− 122y2 + 7y− 4
=
6. Evaluate the limit or, if it does not exist, write dne: limy→−4
y2 + y− 122y2 + 7y− 4
=
In the following three questions, use the definition of the derivative to find the derivative of thefunction.
7. The derivative of s(t) = 3t− 5t2 is given by s′(t) =
8. The derivative of f (x) =√
x− 2 is given by f ′(x) =
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5. Applications of Differentiation
5.1. Tangents & Normals
To start, press Input style:3y2
4√
xis entered as 3/(4x^0.5), etc.
1. The line tangent to y = 3x2 + 4x + 2 at (−1, 1) is y =
2. The line normal to xy + y2 + 2 = 0 at (3,−1) is y =
3. The line normal to s(t) =√
8t + 1 with a slope of −3/4 is y =
4. The line tangent to y =1
(x + 1)2 with a slope of 27/4 is y =
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5.2. Related Rates
To start, press All answers should be numerical. Do not enter units.
1. x and y are both functions of t, related by 5y− x2 = 5.
Ifdxdt
= 3, finddydt
when x = 2:dydt
∣∣∣∣x=2
=
2. A ladder that is 10 m long rests on horizontal ground against a vertical wall.The bottom of the ladder begins to slide away from the wall at 1 m/s.How fast (in m/s) is the top of the ladder sliding down the wall whenthe bottom of the ladder is 6 m away from the bottom of the wall?
3. A spherical exercise ball is inflated at the rate of 0.09 m3/min.At what rate (in mm/s) is the diameter increasing when the diameter is 600 mm?
(The volume of a sphere is V =43
πr3.)
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6. Midterm Practice (Differentiation and Applications)
6.1. Quiz
To start, press
1. Use the product rule to differentiate y = (3− x)(
3 +1x
).
dydx
=
2. Use the quotient rule to differentiate s(t) =2t2 + 1t2 + 1
.dsdt
=
3. g(x) =√
3− x2.dgdx
=
4. Given the implicit function x3 + 3y− y2 = 17:
(a) Differentiate the function with respect to x.dydx
=
(b) Evaluate the slope of the curve at (3, 5).
5. The equation of the line normal to the curve
y = x2 − 3x− 2 at the point (4, 2) is given by y =
6. The elevation of a gun shell is given by y(t) = 300t− 5t2 where y is in metres
and t is in seconds. Find the maximum elevation of the shell (in metres).
7. A spherical weather balloon is released and ascends to the upper atmosphere. As the balloon climbs,atmospheric pressure decreases and the volume of the balloon increases. When it reaches an altitude of12 km, the radius of the balloon is 1.8 m and increases at a rate of 28 mm/min.
At what rate (in m3/min) is the volume of the balloon increasing at this altitude?
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6.2. Quiz
To start, press
1. Use the product rule to differentiate y =(
3− x2)(
3 +1x2
).
dydx
=
2. Use the quotient rule to differentiate s(t) =3− t2
3 + t2 .dsdt
=
3. g(x) =√
1 + 3x2.dgdx
=
4. Given the implicit function x2 − y− y2 + 2 = 0:
(a) Differentiate the function with respect to x.dydx
=
(b) Evaluate the slope of the curve at (2, 3).
5. The equation of the line normal to the curve
y = 3x2 − 4x− 3 at the point (2, 1) is given by y =
6. The daily profit of a small company is given by p(x) = 23x− 0.1x2 where p is in metresand x is the number of products sold.
Find the maximum daily profit for the company (in dollars).
7. A spherical weather balloon is released and ascends to the upper atmosphere. As the balloon climbs,atmospheric pressure decreases and the volume of the balloon increases. When it reaches an altitude of18 km, the radius of the balloon is 3.7 m and increases at a rate of 46 mm/min.
At what rate (in m3/min) is the volume of the balloon increasing at this altitude?
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7. Integration
7.1. Indefinite Integration
To start, press
1.∫
4t3 − 3t2 + 2t− 1 dt = +C
2. If f (x) = x2 − 2x, then∫
3 f (x) dx = +C
3.∫ 1
x3 −1x2 dx = +C
4.∫ √
x3 +3√
x2 dx = +C
5.∫
x(√
x +1√x
)dx = +C
6.∫
(t + 1)2 dt = +C
7.∫ ds
(s + 1)2 = +C
8.∫
3√
p2 − 2p + 1 dp = +C
9.∫
x2√
1− x3 dx = +C
10. The equation of curve with slope 3√
x that
passes through the point (8, 14) is given by y =
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7.2. Definite Integration
To start, press
1.∫ 3
1t3 dt =
2.∫ 3
0
dx√1 + x
=
3.∫ 1
−2s(
1− s2)
ds =
4.∫ 3
−2x2 +
1x2 dx =
5.∫ 14
1
dy3√
4y2 − 4y + 1=
6.∫ 3
0
x3√
x2 − 1dx =
7.∫ 4
0x√
x2 + 9 dx =
8.∫ 5
0(6− 8x)
√2x2 − 3x + 1 dx =
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8. Quiz 3 Practice (Integration)
8.1. Quiz
To start, press
1.∫ 4t4 − 3t3
tdt = +C
2. The slope of the tangent of a curve is given by f (x) = 2x3 − 3x. Given that the curve
passes through the point (2,9), the equation of the curve is given by y =
3.∫ (
y5/3 +1
y3/5
)dy = +C
4.∫ 1
−27− 2x + x3 dx =
5.∫ 14
0
dt3√
4t2 − 4t + 1=
6.∫ 2
−2(s2 + 2s)
√s3 + 3s2 − 4 ds =
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8.2. Quiz
To start, press
1.∫ 4s4 + 2s2
sds = +C
2. The slope of the tangent of a curve is given by h(s) = s3 + s. Given that the curve
passes through the point (-1,0), the equation of the curve is given by y =
3.∫ (
y4/5 − 1y5/4
)dy = +C
4.∫ 2
−1t− 2t3 + 3 dt =
5.∫ 9
0
ds3√
s2 − 2s + 1=
6.∫ 5
0(6− 8x)
√2x2 − 3x + 1 dx =
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8.3. Quiz
To start, press
1.∫ 4t4 + 3t3
t3 dt = +C
2. The slope of the tangent of a curve is given by g(t) = t2 − 4t. Given that the curve
passes through the point (3,-6), the equation of the curve is given by y =
3.∫ (
y3/4 − 1y4/3
)dy = +C
4.∫ 2
−2(s2 + 2s)
√s3 + 3s2 − 4 ds =
5.∫ 6
−3
dx3√
x2 + 4x + 4=
6.∫ 1
0(t2 + 1)
√3t + t3 dt =
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9. Applications of Integration
9.1. Area
To start, press
1. Find the area bounded by y = x2 − 2x− 3 and the line y = 0. A =
2. Find the area bounded by y = x2 and the line y = x. A =
3. Find the area bounded by y = (x− 1)2 − 2 and y = −x2 + 3. A =
4. The function f (x) = x3 − 3x2 − x + 3 intercepts the x-axis at x = −1, x = 1 and x = 3.
Find the area bounded by f (x), the x-axis, x = −1 and x = 3. A =
5. Find the area bounded by y = x2, y = − x2
4+ 5 and y = 2x. A =
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9.2. Work
To start, press
1. An elevator with a loaded weight of 9000 N is supported by cables weighing 192 N/m.When the elevator is at the basement level, there is 37.5 m of cable out between the elevatorand the cable drum at the top of the elevator shaft.
a) How much work is done raising the elevator from the basement to the fifth floor, a distance of 18.0 m?
W = kN ·m (5 significant digits)
b) If three passengers (with a combined weight of 1875 N) get off the elevator at the fifth floor, howmuch work is done raising the elevator the fifth floor to the tenth floor, a distance of 15.0 m?
W = kN ·m (5 significant digits)
2. A bungee-jumper with weight of 720 N steps off a bridge attached to a bungee cord with normal length50 m. The bungee jumper finally comes to rest at a distance 60 m below the bridge. How much work(in kN·m) is done stretching the bungee cord?
W = N ·M (5 significant digits)
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Solutions to Quizzes
I Question 1. Evaluate limx→42
x21
f (x) =x
21is a polynomial of degree 1 and is continuous for every real x
Then the limit can be found by evaluating the f (x) at x = 42
f (42) =4221
= 2
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I Question 2. Evaluate limx→−2
3− 4x
f (x) = 3− 4x is a polynomial of degree 1 and is continuous forevery real xThe limit can be found by evaluating f (x) at x = −2
limx→−2
3− 4x = f (−2) = 3− (−4x) = 11
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I Question 3. Evaluate limx→3
3x√
x2 − 9
f (x) = 3x√
x2 − 9 is defined at x = 0 and for x2 > 9
Thus, f (x) has a domain of (−∞, 3] ∪ 0∪ [3, ∞).
limx→3−
f (x) does not exist because f (x) is not defined in the region to the
left of x = 3.
Consequently, limx→3
f (x) does not exist.
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I Question 4. limx→3
x− 3
f (x) = x2 is continuous for every real x so
limx→2
x2 = f (2) = 4
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I Question 5. limx→−2
x2 − 4x + 2
limx→−2
x2 − 4x + 2
= limx→−2
(x− 2)(x + 2)(x + 2)
= limx→−2
(x− 2)
= −4
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I Question 6. limx→2
x2 − 4x + 2
f (x) is continuous everywhere except at x = −2. In particular, f (x) is continuous at x = 2
limx→2
x2 − 4x + 2
= f (2) =04= 0
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I Question 7. limx→2
∣∣x− 2∣∣
x− 2
limx→2−
∣∣x− 2∣∣
x− 2= lim
x→2−2− xx− 2
= limx→2−
−1 = −1
limx→2+
∣∣x− 2∣∣
x− 2= lim
x→2+x− 2x− 2
= limx→2+
1 = 1
x
f (x) =
∣∣x− 2∣∣
x− 2
(2,−1)
(2, 1)
Both one-sided limits exist but they are not equal to each other:
limx→2
∣∣x− 2∣∣
x− 2does not exist (d.n.e.)
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I Question 8. limx→9
√x− 3
9− x
limx→9
√x− 3
9− x= lim
x→9
√x− 3(
3−√
x)(
3 +√
x)
= limx→9
−(3−√
x)(
3−√
x)(
3 +√
x)
= limx→9
−13 +√
x
= −16
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I Question 9. Evaluate limx→2
(x− 2)√
x2 − 11
f (x) = (x− 2)√
x2 − 11 is defined at x = 2 since f (2) = 0 ·√−1 = 0.
However, f (x) is not defined as x → 2 since, when x is close to but not equal to 2,x− 2 6= 0.
Then, f (x) equals some small number multiplied by the square root of a negativenumber. The square root of a negative number is not defined(as far as we’re concerned in this course).
So, f (x) is not defined as x → 2 and there is no limit.
limx→2
(x− 2)√
x2 − 11 does not exist.
x
f (x) f (x) = (x− 2)√
x2 − 11
(2, 0) (√
11, 0)
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I Question 10. Evaluate limx→2
x3 − 8x2 − 4
The function f (x) =x3 − 8x2 − 4
is not continuous at x = 2; the denominator approaches
0 as x approaches 2
limx→2
x3 − 8x2 − 4
= limx→2
(x− 2
)(x2 + 2x + 4
)(x− 2
)(x + 2
)= lim
x→2
(x2 + 2x + 4
)(x + 2
)Now, the function g(x) =
(x2 + 2x + 4
)(x + 2
) is well-behaved, and continuous, at x = 2 so
the limit can be easily evaluated:
limx→2
(x2 + 2x + 4
)(x + 2
) = limx→2
g(x) = g(2) = 3
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I Question 1. Evaluate limx→4
√x2 + 9x + 3
f (x) =
√x2 + 9x + 3
is defined and continuous except at x = −3
limx→4
f (x) = f (4) =
√42 + 94 + 3
=57
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I Question 2. Evaluate limx→2
(x− 3)3 +4
2x2
f (x) = (x− 3)3 +4
2x2 is defined and continuous except at x = 0
limx→2
f (x) = f (2) = (2− 3)3 +4
2(22) = −1 +
12= −1
2
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I Question 3. Evaluate limt→−1
3t2 − 4t− 1
s(t) =3t2 − 4t− 1
is defined and continuous except at t = 1
limt→−1
s(t) = s(−1) =3(−1)2 − 4−1− 1
=12
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I Question 4. Evaluate limx→2
x2 + x− 6x2 − x− 2
f (x) =x2 + x− 6x2 − x− 2
limx→2
f (x) is of the form00
; it is indeterminate.
Both the numerator and the denominator are 0 when x = 2; this tells usthat x− 2 is a factor of them both.
limx→2
x2 + x− 6x2 − x− 2
= limx→2
(x + 3)(x− 2)(x + 1)(x− 2)
= limx→2
(x + 3)(x + 1)
=53
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I Question 5. Evaluate limx→∞
1x2
As can be seen from the graph, as x → ∞ the value of f (x) getscloser and closer to 0 but never actually reaches it, i.e. we cannot
find an x large enough that1x2 equals 0.
We can, though, make f (x) as close to 0 as we want by appropriatechoice of x.
Consider an extremely small value d > 0. We need to find an x sothat f (x) ≤ d. Try x = 1/
√d. Then f (1/
√d) = d as we required.
x
f (x) =1x2
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I Question 6. Evaluate limx→2
√4− x2
Consider the domain of f (x) =√
4− x2.
The value under the square root sign cannot be negative. Thus,
4− x2 ≥ 0 ⇒ x2 ≤ 4 ⇒ −2 ≤ x ≤ 2
Since f (x) is not defined for x > 2, limx→2+
does not exist.
Thus, lim[→x
]2 does not exist since one of its one-sided limits does not exist.
Note, however, that limx→2−
f (x) does exist and is equal to 0.
x
f (x) =√
4− x2
(−2, 0) (2, 0)
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I Question 7. Evaluate limx→2
1x −
12
x− 2
Division by x− 2 is legal in the following calculations because althoughx → 2, x never actually reaches 2 and x 6= 2 so there is no division by 0.
limx→2
1x −
12
x− 2= lim
x→2
2−x2x
x− 2
= − limx→2
x−22x
x− 2
= − limx→2
12x1
= − limx→2
1x
= −14
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I Question 8. limx→2
x3 − 8x− 2
Althoughx3 − 8x− 2
is not defined for x = 2,
it is defined on each side of x = 2.Therefore,
limx→2
x3 − 8x− 2
= limx→2
(x− 2)(
x2 + 2x + 4)
(x− 2)
= limx→2
(x2 + 2x + 4
)
=((2)2 + 2(2) + 4
)= 12
x
f (x)f (x) =
x3 − 8x− 2
(2, 12)
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I Question 9. limx→−1
x3 + 1x + 1
Althoughx3 + 1x + 1
is not defined for x = −1,
it is defined on each side of x = −1.Therefore,
limx→−1
x3 + 1x + 1
= limx→−1
(x + 1)(
x2 − x + 1)
(x + 1)
= limx→−1
(x2 − x + 1
)
=((−1)2 − (−1) + 1
)= 3
x
f (x)
f (x) =x3 + 1x + 1(−1, 3)
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I Question 10. limx→−1
x3 + 1x + 1
Both the numerator x2 + 3t− 10 and the denominator x2 − 4 approach 0as x → 2 so the function is not continuous at x = 2. We use the fact thatx → 2 but x 6= 2 to divide both numerator and denominator by x− 2 toget a continuous function:
limx→2
x2 + 3t− 10x2 − 4
= limx→2
(x + 5)(x− 2)(x + 2)(x− 2)
= limx→2
(x + 5)(x + 2)
=(2 + 5)(2 + 2)
= 74
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I Question 1. f (x) = 3x− x2.
f ′(x) = limh→0
f (x + h)− f (x)h
= limh→0
3(x + h)− (x + h)2 −(3x− x2)
h
= limh→0
3x + 3h− x2 − 2hx− h2 − 3x + x2
h
= limh→0
3h− 2hx− h2
h= lim
h→03− 2x− h
= 3− 2x
Solution should be input 3-2x or equivalent
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I Question 2. 3t2 + 2t + 1
f ′(t) = limh→0
f (t + h)− f (t)h
= limh→0
(t + h)3 + (t + h)2 + (t + h)−(t3 + t2 + t
)h
= limh→0
t3 + 3t2h + 3th2 + h3 + t2 + 2th + h2 + t + h− t3 − t2 − th
= limh→0
3t2h + 3th2 + h3 + 2th + h2 + hh
= limh→0
3t2 + 3th + h2 + 2t + h + 1
= 3t2 + 2t + 1
Solution should be input 3t^2+2t+1 or equivalent
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I Question 3. f (x) = 2x +2x
.
f ′(x) = limh→0
f (x + h)− f (x)h
= limh→0
2(x + h) + 2x+h − 2x− 2
xh
= limh→0
2x + 2h + 2x+h − 2x− 2
xh
= limh→0
2h + 2x−2(x+h)x(x+h)
h
= limh→0
2h− 2hx(x+h)
h
= limh→0
2− 2x(x + h)
= 2− 2x2
Solution should be input 2-2/x^2 or equivalent
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I Question 4. f (s) =s
2s + 1..
f ′(s) = limh→0
f (s + h)− f (s)h
= limh→0
s+h2(s+h)+1 −
s2s+1
h
= limh→0
(s+h)(2s+1)−s(2(
s+h)+1)
(2(
s+h)+1)(2s+1)
h
= limh→0
2s2+s+2sh+h−2s2−2hs−s(2(
s+h)+1)(2s+1)
h
= limh→0
h(2(
s+h)+1)(2s+1)
h
= limh→0
1(2(s + h
)+ 1)(2s + 1)
=1
(2s + 1)2
Solution should be input 1/(2s+1)^2 or equivalent
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I Question 5. f (x) =√
x + 1.
f ′(x) = limh→0
f (x + h)− f (x)h
= limh→0
√x + h + 1−
√x + 1
h
= limh→0
√x + h + 1−
√x + 1
h·√
x + h + 1 +√
x + 1√x + h + 1 +
√x + 1
= limh→0
1h· (x + h + 1)− (x + 1)√
x + h + 1 +√
x + 1
= limh→0
1h· h√
x + h + 1 +√
x + 1
= limh→0
1√x + h + 1 +
√x + 1
=1
2√
x + 1
Solution should be input 1/(2(x+1)^0.5) or equivalent
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I Question 6. s(t) = 42t− 5t2. Find the value of t where the tangent to s(t) is horizontal.
s′(t) = limh→0
s(t + h)− s(t)h
= limh→0
42(t + h)− 5(t + h)2 − 42t + 5t2
h
= limh→0
42t + 42h− 5t2 − 10th− 5h2 − 42t + 5t2
h
= limh→0
42h− 10th− 5h2
h= lim
h→042− 10t− 5h
= 42− 10t
Then 42− 10t = 0→ t = 4.2
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I Question 7. f (x) =√
4− x2. Find x such that f ′(x) = −1.
f ′(x) = limh→0
f (x + h)− f (x)h
= limh→0
√4− (x + h)2 −
√4− x2
h
= limh→0
√4− (x + h)2 −
√4− x2
h·√
4− (x + h)2 +√
4− x2√4− (x + h)2 +
√4− x2
= limh→0
1h·(4− (x + h)2)− (4− x2)√
4− (x + h)2 +√
4− x2
= limh→0
1h· x2 − (x + h)2√
4− (x + h)2 +√
4− x2
= limh→0
1h· −h(2x + h)√
4− (x + h)2 +√
4− x2
= limh→0
−(2x + h)√4− (x + h)2 +
√4− x2
=−(2x)
2√
4− x2=
−x√4− x2
Then:
−x√4− x2
= −1
⇒ x =√
4− x2
⇒ x2 = 4− x2
⇒ 2x2 = 4
⇒ x =√
2
Solution should be input 2^0.5 or equivalent
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I Question 8.d
dx
(1√x
).
f (x + h)− f (x) =1√
x + h− 1√
x
=
√x−√
x + h√
x√
x + h
=
√x−√
x + h√
x√
x + h·√
x +√
x + h√
x +√
x + h
=x− (x + h)
√x√
x + h(√
x +√
x + h)
f (x + h)− f (x)h
=−h
h√
x√
x + h(√
x +√
x + h)
ddt( 1√
t
)= lim
h→0
−1√
x√
x + h(√
x +√
x + h)
=−1
√x√
x + h(√
x +√
x + h)
= − 1√x√
x(√
x +√
x)
= − 12x√
x
= −12
x−32
Solution should be input 1/(2x^(3/2)) or equivalent
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I Question 8. The tangent line at
(4,
12
)
The tangent line at
(4,
12
):
ddx
(1√
t
)=−1
2x√
xd
dx
(1√
t
)x=4
=−1
2× 4√
4= − 1
16
Equation of the line:
y− 12
x− 4= − 1
16
y = − x + 416
(or − x
16− 1
4, if you prefer
)
f (x)
f (x) =1√x
f (x) = − (12− x)16
Solution should be input -(x+4)/16 or equivalent
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I Question 9. f (x) =1− x1 + x
.
f (x + h)− f (x) =1− (x + h)1 + (x + h)
− 1− x1 + x
=
[1− (x + h)
](1 + x)− (1− x)
[1 + (x + h)
](1 + (x + h)
)(1 + x
)=
1 + x− x− x2 − h− hx− 1− x− h + x + x2 + xh(1 + (x + h)
)(1 + x
)=
−2h(1 + (x + h)
)(1 + x
)
f ′(x) = limh→0
f (x + h)− f (x)h
= limh→0
−2hh(1 + (x + h)
)(1 + x
)= lim
h→0
−2(1 + (x + h)
)(1 + x
)f ′(x) =
−2(1 + x)2
Solution should be input -2/(1+x)^2) or equivalent
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I Question 1. f (x) = 4x3 − 3x2 + 2x− 1.
f ′(x) =d
dx4x3 − d
dx3x2 +
ddx
2x− 1
= 12x2 − 6x + 2
Solution should be input 12x^2-6x+2 or equivalent
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I Question 2. s(t) = (3t)4 − (2t)5
s′(t) =d
dx(3t)4 − d
dx(2t)5
= 34 ddx
t4 − 25 ddx
t5
= 34 × 4t3 − 25 × 5x4
= 324t3 − 160t4
Solution should be input 324t^3 - 160t^4 or equivalent
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I Question 3. f (x) =√
2x
f ′(x) =d
dx
√2x
=√
2d
dx√
x
=√
2d
dxx1/2
=√
2 · 12
x−1/2
=
√2
2√
x
=1√2x
Solution should be input (2x)^(-1/2) or equivalent
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I Question 4. g(t) = π(√
4t2 − 3√
8t3)(√
4t2 +3√
8t3)
√4t2 = 2t and
3√
8t3 = 2t so:
g(t) = π(2t− 2t)
(2t + 2t) = 0
g′(t) = 0
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I Question 5. f (x) =(1− x3)(x3 + 1
)
f (x) = 1− x6
f ′(x) = −6x5
Solution should be input -6x^5 or equivalent
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I Question 6. f (x) = x2√x− 1√x+ x3√
x3
f (x) = x5/2 − x−1/2 + x9/2
f ′(x) =52
x3/2 −(−1
2x−3/2
)+
92
x7/2
=5x√
x2
+1
2x√
x+
9x3√x2
Solution should be input 5/2*x^(3/2)+1/(2x^(3/2))+9/2*x^(7/2) or equivalent
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I Question 7. h(t) = 22t− 5t2, find h′(1.5)
h′(t) = 22− 10t
h′(1.5) = 7
After 1.5 seconds, the ball is travelling upwards at 7 m/s.
Solution should be input 7. Do not enter units.
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I Question 7. h(t) = 22t− 5t2, solve h′(t) = 0
h′(t) = 22− 10t
h′(t) = 0⇒ t = 2.2
The ball reaches its highest point after 2.2 seconds.
Solution should be input 2.2. Do not enter units.
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I Question 7. h(t) = 22t− 5t2. Find maximum value for h.
h(2.2) = 22(2.2)− 5(2.2)2
= 24.2
The ball reaches its highest point after 2.2 seconds.
Solution should be input 24.2. Do not enter units.
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I Question 1. h(t) = (2t + 3)(4t− 5). Find h′(t)
Let u(t) = (2t + 3) and v(t) = 4t− 5).Then h(t)) = u(t)v(t) and
h′(t) = udvdt
+ vdudt
= (2t + 3)4 + (4t− 5)2
= 8t + 12 + 8t− 10
= 16t + 2
Solution should be input 16t+2 or equivalent.
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I Question 2. y(x) =(2x3 − 3
)(3x2 + 2
). Find y′(x).
Let u(x) =(2x3 − 3
)and v(x) =
(3x2 + 2
).
Then y(x) = u(x)v(x) and
y′(x) = (uv)′ = uv′ + vu′
=(2x3 − 3
)(6x) +
(3x2 + 2
)(6x2)
= 12x4 − 18x + 18x4 + 12x2
= 30x4 + 12x2 − 18x
Solution should be input 30x^4+12x^2-18x or equivalent.
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I Question 3. g(x) =(3x2 + 2x + 1
)2. Find f ′(x)
Let u(x) =(3x2 + 2x + 1
)and v(x) =
(3x2 + 2x + 1
).
Then g(x) = u(x)v(x) and
g′(x) = (uv)′ = uv′ + vu′
=(3x2 + 2x + 1
)(6x + 2) +
(3x2 + 2x + 1
)(6x + 2)
= 2(3x2 + 2x + 1
)(6x + 2)
= 4(3x2 + 2x + 1
)(3x + 1)
Solution should be input 4(3x^2+2x+1)(3x+1) or equivalent.
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I Question 4. Evaluatedds(s2 + s + 1
)(s2 − s + 1
)
Let u(s) =(s2 + s + 1
)and v(s) =
(s2 − s + 1
).
Then f (s) = u(s)v(s) and
f ′(s) =dds
(uv) = udvds
+ vduds
=(s2 + s + 1
)(2s− 1
)+(s2 − s + 1
)(2s + 1
)= 2s3 + 2s2 + 2s− s2 − s− 1 + 2s3 − 2s2 + 2s + s2 − s + 1
= 4s3 + 2s
Solution should be input 4s^3+2s or equivalent.
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I Question 5. y =
(2√
x +3x
)(3√
x− 2x
)Find y′
Let u(x) =(
x2 + x + 1)
and v(x) =(
x2 − x + 1).
Then g(x) = u(x)v(x) and
y =
(2√
x +3x
)(3√
x− 2x
)
y′ =(
2√
x +3x
)(3√
x− 2x
)′+
(2√
x +3x
)′ (3√
x− 2x
)
=(
2x1/2 + 3x−1) (
3x1/2 − 2x−1)′
+(
2x1/2 + 3x−1)′ (
3x1/2 − 2x−1)
=(
2x1/2 + 3x−1)(3
2x−1/2 + 2x−2
)+(
x−1/2 − 3x−2) (
3x1/2 − 2x−1)
=
(2x1/2 · 3
2x−1/2 + 2x1/2 · 2x−2 + 3x−1 · 3
2x−1/2 + 3x−1 · 2x−2
)+(
x−1/2 · 3x1/2 − x−1/2 · 2x−1 − 3x−2 · 3x1/2 + 3x−2 · 2x−1)
=
(3 + 4x−3/2 +
92
x−3/2 + 6x−3)+(
3− 2x−3/2 − 9x−3/2 + 6x−3)
= 6− 52x3/2 +
12x3
Solution should be input 6-5/(2x^(3/2))+12/x^3 or equivalent.
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I Question 6. y(x) =x− 3x + 3
. Find y′(x)
Let u(x) = x− 3 and v(x) = x + 3.
Then y(x) =u(x)v(x)
and
y′(x) =
(u(x)v(x)
)′=
v(x)u′(x)− u(x)v′(x)v2(x)
=(x + 3)(x− 3)′ − (x− 3)(x + 3)′
(x + 3)2
=(x + 3) · 1− (x− 3) · 1
(x + 3)2
=(x + 3)− (x− 3)
(x + 3)2
=6
(x + 3)2
Solution should be input 6/(x+3)^2 or equivalent.
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I Question 7. g(t) =t2
t + 1. Find g′(t).
Let u(t) = t2 and v(t) = t + 1.
Then g(t) =u(t)v(t)
and
(uv
)′=
vu′ − uv′
v2
=(t + 1) · 2t− t2 · 1
(t + 1)2
=2t + 2t2 − x2
(t + 1)2
=t2 + 2t(t + 1)2
=t(t + 2)(t + 1)2
Solution should be input t(t+2)/(t+1)^2 or equivalent.
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I Question 8. y(x) =x2 + 1
2x2 + 1. Find y′(x)
y′(x) =
(2x2 + 1
)(2x)−
(x2 + 1
)(4x)(
2x2 + 1)2
=4x3 + 2x− 4x3 − 4x(
2x2 + 1)2
=−2x(
2x2 + 1)2
Solution should be input -2x/(2x^2+1)^2 or equivalent.
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I Question 9. p(s) =s2 + 1√
x− 1. Find p′(s)
p′(s) =(√
s− 1)(2s)−(s2 + 1
) ( 12√
s
)(√
s− 1)2
=2s3/2 − 2s− 1
2
(s3/2 + s−1/2
)(√
s− 1)2
=4s3/2 − 4s−
(s3/2 + s−1/2
)2(√
s− 1)2
=3s3/2 − 4s− s−1/2
2(√
s− 1)2
=3s√
s− 4s− 1√s
2(√
s− 1)2
Solution should be input ( 3s^(3/2)-4s-1/s^(1/2))/2/(s^(1/2)-1)^2 or equivalent.
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I Question 10. y(x) =1
1 + 1x
. Find y′.
y′ =
(1 + 1
x
)(1)′ − (1)
(1 + x−1)′(
1 + 1x
)2
=
(1 + 1
x
)(0)−
(1 + x−1)′(
1 + 1x
)2
=−x−2(1 + 1
x
)2
=−1
x2(
1 + 1x
)2
=−1
(x + 1)2
Note:1
1 + 1x=
xx + 1
which would have been an easier differentiation problem!
Solution should be input -1/(x+1)^2 or equivalent.
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I Question 1. f (x) = (3x + 2)3. Find f ′(x).
Let u = 3x + 2. Then f (u) = u3 andd fdu
= 3u2
Also,dudx
= 3 so:
f ′(x) =d fdx
=d fdu· du
dx= 3u2 · 3 = 9u2 = 9(3x + 2)2
Solution should be input 9(3x+2)^2 or equivalent.
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I Question 2. g(t) = (10− 3t2)5. Find g′(t)
Let u = 10− 3t2. Then g(u) = u5 anddgdu
= 5u4
Also,dudt
= −6t so:
g′(t) =dgdt
=dgdu· du
dt= 5u4(−6t) = −30tu4 = −30t
(10− 3t2
)4
Solution should be input -30t(10-3t^2)^4 or equivalent.
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I Question 3. h(x) =1
(x2 + 2)2 . Find h′(x).
Let u = x2 + 2. Then h(u) = u−2 anddhdu
= −2u−3
Also,dudx
= 2x so:
h′(x) =dhdu· du
dx= −2u−3(2x) = −4x
(x2 + 2
)−3=
−4x
(x2 + 2)3
Solution should be input -4x/(x^2+2)^3 or equivalent.
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I Question 4. y =√
3x2 − 2x + 1. Finddydx
Let u = 3x2 − 2x + 1. Then y(u) = u12 ⇒ dy
du=
12
u−12
Also,dudx
= 6x− 2 so:
y′(x) =12
u−12 (6x− 2) = (3x− 1)
(3x2 − 2x + 1
)− 12=
3x− 1√3x2 − 2x + 1
Solution should be input (3x-1)/(3x^2-2x+1)^0.5 or equivalent.
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I Question 5. y =(
1 + x3)1/3
. Finddydx
.
By now you should be able to recognise the substitution for u and applythe chain rule “on the fly”.
y =(
1 + x3)1/3
y′ =13
(1 + x3
)−2/3· 3x2
=x2
3√(1 + x3)
2
Solution should be input (3x-1)/(3x^2-2x+1)^0.5 or equivalent.
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I Question 6. s(t) =
√5t
. Find s′(t).
s =
√5t=√
5 · t−1/2
s′ =√
5(−1
2t−3/2
)= −1
2
√5t3
= − 12t
√5t
Solution should be input -1/2*(5/t^3)^(1/2) or equivalent.
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I Question 7. y = x3(
2x2 − 3)4
. Find y′.
y = x3(
2x2 − 3)4
y′ = x3 · ddx
[(2x2 − 3
)4]+(
2x2 − 3)4· d
dx
[x3]
= x3[
4(
2x2 − 3)3· 4x]+(
2x2 − 3)4 [
3x2]
= 16x4(
2x2 − 3)3
+ 3x2(
2x2 − 3)4
= x2(
2x2 − 3)3 [
16x2 + 3(
2x2 − 3)]
= x2(
2x2 − 3)3 (
22x2 − 9)
Solution should be input x^2(2x^2-3)^3(22x^2-9) or equivalent.
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I Question 8. y = (2x + 1)3(3x + 1)2. Find y′.
y = (2x + 1)3(3x + 1)2
y′ = (2x + 1)3 · ddx
[(3x + 1)2
]+ (3x + 1)2 · d
dx
[(2x + 1)3
]= (2x + 1)3 · [2(3x + 1) · 3] + (3x + 1)2 ·
[3(2x + 1)2 · 2
]= (2x + 1)3 [6(3x + 1)] + (3x + 1)2 [6(2x + 1)]
= 6(2x + 1)(3x + 1)[(2x + 1)2 + (3x + 1)
]= 6(2x + 1)(3x + 1)
[2x2 + 4x + 1 + 3x + 1)
]= 6(2x + 1)(3x + 1)
[2x2 + 7x + 2)
]
Solution should be input 6(2x+1)(3x+1)(2x^2+7x+2) or equivalent.
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I Question 9. s(t) =(
1 + t2/3)3/2
. Finddsdt
.
(This problem requires a repeated application of the chain rule.)
s =(
1 + t2/3)3/2
dsdt
=32
(1 + t2/3
)1/2· d
dt
(1 + t2/3
)=
32
(1 + t2/3
)1/2· 2
3t1/3
= t1/3√
1 + t2/3
Solution should be input t^(1/3)*(1+t^(2/3))^(1/2) or equivalent.
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I Question 10. g(y) =[1 + (2y + 3)4
]5. Find g′(y).
(This problem requires a repeated application of the chain rule.)
g(y) =[1 + (2y + 3)4
]5
g′(y) = 5[1 + (2y + 3)4
]4· d
dy
[1 + (2y + 3)4
]= 5
[1 + (2y + 3)4
]4·[4(2y + 3)3 · 2
]= 40
[1 + (2y + 3)4
]4(2y + 3)3
Solution should be input 40(1+(2y+3)^4)^4(2y+3)^3 or equivalent.
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I Question 1.
x + xy + y = 0. Find y′
ddx(
x)+
ddx(xy)+
ddx(y)=
ddx(0)
1 + (xy′ + y) + y′ = 0
y′(x + 1) = −(y + 1)
y′ = − y + 1x + 1
Solution should be input -(y+1)/(x+1) or equivalent.
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I Question 1.
y′ = − y + 1x + 1
. Find y′(−2,−2), y′(0, 0) and y′(2,−2/3)
y′(−2,−2) = −−2 + 1−2 + 1
= −1
y′(0, 0) = −0 + 10 + 1
= −1
y′(2,−2/3) = −−2/3 + 12 + 1
= −−1/33
= −1/9
Solution should be input -1, -1, -1/9 or decimal equivalent.
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I Question 1.
x + xy + y = 0. Find y in terms of x:
x + xy + y = 0
x + y(x + 1) = 0
y(x + 1) = −x
y = g(x) =−x
x + 1
Solution should be input -x/(x+1) or equivalent.
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I Question 1.
g(x) = − xx + 1
. Finddgdx
g(x) = − xx + 1
dgdx
= − vu′ − uv′
v2
= − (x + 1)(1)− x(1)(1 + x)2
= − 1(1 + x)2
Solution should be input -1/(x+1)^2 or equivalent.
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I Question 1.
g′(x) = − 1(x + 1)2 . Find g′(−2), g′(0) and g′(2)
g′(−2) = − 1(−2 + 1)2 = −1
g′(0) = − 1(0 + 1)2 = −1
g′(2) = − 1(2 + 1)2 = −1
9=
Solution should be input -1, -1, -1/9 or decimal equivalent.
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I Question 2.
f (x, y) = x2y2 − x3 + y3 = 0. Determinedydx
Differentiate with respect to x:
2x + 2ydydx
= 3x2 + 3y2 dydx
dydx(2y− 3y2) = 3x2 − 2x
dydx
=x(3x− 2
)y(y− 3y
)Solution should be input (3x^2-2x)/(2y-3y^2) or equivalent.
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I Question 3.
x2y2 = x3 + y3. Determinedydx
Differentiate with respect to x:
x2 · 2yy′ + y2 · 2x− 3x2 + 3y2y′ = 0
y′(
2x2y + 3y2)= 3x2 − 2xy2
y′ =x(3x− 2y2)
y (3y + 2x2)
Solution should be input (3x^2-2xy^2)/(2x^2y+3y^2) or equivalent.
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I Question 4.
a) (x− 2)2 + (y− 1)2 − 25 = 0 Finddydx
anddydx
∣∣∣∣(x,y)=(5,5)
Differentiate with respect to x:
(x− 2)2 + (y− 1)2 = 25
x2 − 4x + 4 + y2 − 2y + 1 = 25
x2 − 4x + y2 − 2y = 20
2x− 4 + 2ydydx− 2
dydx
= 0
dydx
(2y− 2) = 4− 2x
dydx
=4− 2x2y− 2
=2− xy− 1
Solution should be input (x-2)/(1-y) or equivalent.
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I Question 5. (x + y)2 + (x− y)2 = x3 + y3. Finddydx
Expand, then differentiate with respect to x:
x2 + 2xy + y2 + x2 − 2xy+2 = x3 + y3
2x2 + 2y2 = x3 + y3
4x + 4ydydx
= 3x2 + 3y2 dydx
dydx(4y− 3y2) = 3x2 − 4x
dydx
=x(3x− 4)y(4− 3y)
Alternatively, without expanding the quadratics:
2(x + y)(
1 +dydx
)+ 2(x + y)
(1− dy
dx
)= 3x2 + 3y2 dy
dx
2[(
x + xdydx
+ y + ydydx
)+
(x− x
dydx− y + y
dydx
)]= 3x2 + 3y2 dy
dx
4x + 4ydydx
= 3x2 + 3y2 dydx
dydx(4y− 3y2) = 3x2 − 4x
dydx
=x(3x− 4)y(4− 3y)
Solution should be input x(3x-4)/y/(4-3y) or equivalent.
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I Question 6.
x2y− xy2 + x2 + y2 = 0. Find y′
ddx(
x2y)− d
dx(
xy2)+ ddx(
x2)+ ddx(y2) = d
dx(0)(
2xy + x2y′)−(
x 2yy′ + y2(1))+ 2x + 2yy′ = 0
y′(
x2 − 2xy + 2y)+(2xy− y2 + 2x
)= 0
y′ =y2 − 2x− 2xyx2 + 2y− 2xy
Solution should be input (y^2-2x-2xy)/(x^2+2y-2xy) or equivalent.
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I Question 1. Evaluate limx→2
2x2 − x + 3
The function 2x2 − x + 3 is a polynomial and continuous everywhere.
The limit is found by evaluating the function at x = 2:
limx→2
2x2 − x + 3 = 2(2)2 − (2) + 3
= 9
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I Question 2. Evaluate limt→42
√t + 7
The function√
t + 7 is continuous in the region surrounding t = 42.
The limit is found by evaluating the function at t = 42:
limt→42
√t + 7 =
√42 + 7
=√
49
= 7
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I Question 3. Evaluate limx→−1
2− xx2 + 1
The function f (x) =2− xx2 + 1
is continuous for every real value x since
the denominator can never equal 0.
The limit is found by evaluating the function at x = −1:
limx→−1
2− xx2 + 1
=2− (−1)(−1)2 + 1
=32
= 1.5
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I Question 4. Evaluate limh→√
3
√h2 − 3
This limit does not exist because the left limit does not exist.
The function f (h) =√
h2 − 3 is not defined for h <√
3 since the value under the radicalsign would then be negative and there is no real number with a negative square root.
(The right limit does exist and is equal to 0.)
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I Question 5. Evaluate limx→−2/3
6 + 9x|3x + 2|
This limit does not exist! Both one-sided limits do exist but they are not equal.
If x > 2/3, then |3x + 2| = 3x + 2 and6 + 9x|3x + 2| =
6 + 9x3x + 2
=3(3x + 2)
3x + 2= 3
If x < 2/3, then |3x + 2| = −3x− 2 and6 + 9x|3x + 2| =
6 + 9x−3x− 2
=3(3x + 2)−(3x + 2)
= −3
(If x = 2/3, then6 + 9x|3x + 2| is not defined.)
x
f (x) =6 + 9x∣∣3x + 2
∣∣
(−2/3,−3)
(−2/3, 3)
limx→−2/3−
6 + 9x|3x + 2| = −3
limx→−2/3+
6 + 9x|3x + 2| = 3
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I Question 6. Evaluate lims→−1
1 + 3s + s2
s
This function is not continuous at s = 0 since the denominator of thefunction is then 0, but the function is continuous for every s 6= 0.Thus, the limit as s→ −1 can be found by evaluation of the functionat s = −1:
lims→−1
1 + 3s + s2
s=
1 + 3(−1) + (−1)2
−1
= lims→−1
(x + 1)
=−1−1
= 1
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I Question 7. Evaluate limt→5
t2 − 4t− 5t− 5
Both the numerator and the denominator of the limit approach 0 ast→ 5 so some algebra is necessary:
• Polynomial functions are continous everywhere.
• Since both g(t) = t2 − 4t− 5 and h(t) = t− 5 are polynomialfunctions that approach 0 as t→ 5, it follows that g(5) = 0and h(5) = 0.
• Thus, t− 5 is a factor of both g(t) and h(t).
• Factor g(t) and h(t), then cancel the (t− 5) factor.
• This leaves a function f (x) =g(t)h(t)
= x + 1 that is continuous
at t = 5.
limt→5
t2 − 4t− 5t− 5
= limt→−5
(t− 5)(t + 1)(t− 5)
= limx→5
(x + 1)
= 6
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I Question 8. Find the derivative of s(t) = 3t− 5t2 using the definition of the derivative:
s′(t) = limh→0
s(t + h)− s(t)h
= limh→0
3(t + h)− 5(t + h)2 −[3t− 5t2]
h
= limh→0
�3t + 3h−��5t2 − 10th− 5h2 −�3t +��5t2
h
= limh→0
3− 10t− 5h
= 3− 10t
Solution should be input 3-10t or equivalent
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I Question 9. Find the derivative of f (x) =√
x− 2 using the definition of the derivative:
f ′(x) = limh→0
f (x + h)− f (x)h
= limh→0
√(x + h)− 2−
√x− 2
h
= limh→0
√(x + h)− 2−
√x− 2
h·√(x + h)− 2 +
√x− 2√
(x + h)− 2 +√
x− 2
= limh→0
1h· (x + h− 2)− (x− 2)√
x + h− 2 +√
x− 2
= limh→0
1h· h√
x + h− 2 +√
x− 2
= limh→0
1√x + h− 2 +
√x− 2
=1
2√
x− 2
Solution should be input 2/(x-2)^0.5 or equivalent
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I Question 1. Evaluate limx→2
2(x− 1)2 − 2(x + 1)
The function contains only terms in x2, x and constants; it is a polynomial.Polynomials are continuous for all real numbers x so the limit is found byevaluating the function at x = 2
limx→2
2(x− 1)2 − 2(x + 1) = 2(2− 1)2 − 2(2 + 1) = −4
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I Question 2. Evaluate limx→1
√x2 + 4x + 2
The function f (x) = limx→1
√x2 + 4x + 2
is continuous for every real value
x 6= −2. In particular, it is continuous at x = 1.
The limit is found by direct evaluation of the function at x = 1:
limx→1
√x2 + 4x + 2
=
√(1)2 + 4(1) + 2
=
√5
3
Solution should be input 5^0.5/3 or equivalent
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I Question 3. Evaluate lims→2
4− s2
2− s
Both the numerator and the denominator of the limit approach 0 ast→ −5 so some algebra is necessary:
• Polynomial functions are continous everywhere.
• Since both g(t) = t2 − 4t− 5 and h(t) = t− 5 are polynomialfunctions that approach 0 as t→ 5, it follows that g(5) = 0and h(5) = 0.
• Thus, t− 5 is a factor of both g(t) and h(t).
• Factor g(t) and h(t), then cancel the (t− 5) factor.
• This leaves a function f (x) =g(t)h(t)
= x + 1 that is continuous
at t = 5.
lims→2
4− s2
2− s= lim
s→1
(2− s)(2 + s)2− s
= lims→1
(2 + s)
= 4
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I Question 4. Evaluate limx→2
x2 + 5x− 14x2 + x− 6
Note: A function of the type r(x) = p(x)/qx, where both p(x) and q(x) are polynomialfunctions is called a rational function.
As in the previous question, both the numerator and the denominator of the rational
function r(x) =x2 + 5x− 14
x2 + x− 6are 0 at x = 2, so x− 2 is a factor of both the numerator
and the denominator.
limx→2
x2 + 5x− 14x2 + x− 6
= limx→2
(x− 2)(x + 7)(x− 2)(x + 3)
= limx→2
(x + 7)(x + 3)
=95
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I Question 5. Evaluate limx→1
∣∣x2 − 1∣∣
x− 1
This limit does not exist! Both one-sided limits do exist but they are not equal.
x
f (x) f (x) =
∣∣x2 − 1∣∣
x− 1
(1,−2)
(1, 2)
limx→1+
∣∣x2 − 1∣∣
x− 1= lim
x→1+x2 − 1x− 1
= limx→1+
(x− 1)(x + 1)(x− 1)
= limx→1+
(x + 1) = 2
limx→1−
∣∣x2 − 1∣∣
x− 1= lim
x→1−1− x2
x− 1= lim
x→1+− x2 − 1
x− 1= lim
x→1+−(x + 1) = −2
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I Question 6.
This limit does not exist because the left-limit does not exist;√(t− 2)3
t− 2is undefined for x < 2.
limt→2
√(t− 2)3
t− 2= lim
t→2
(t− 2)√(t− 2)
t− 2
= limt→2
√(t− 2)
limt→2+
√(t− 2) = 0
limt→2−
√(t− 2) does not exist - square root of a negative number!
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I Question 7. Find the derivative of y(x) =1
1 + xusing the definition of the derivative:
y′(t) = limh→0
y(x + h)− y(x)h
= limh→0
[1
x + h + 1− 1
x + 1
]· 1
h
= limh→0
x + 1− (x + h + 1)(x + h + 1)(x + 1)
· 1h
= limh→0
�x + �1− �x− h− �1(x + h + 1)(x + 1)
· 1h
= limh→0
−1(x + h + 1)(x + 1)
= − 1(x + 1)2
Solution should be input -1/(x+1)^2 or equivalent
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I Question 8.d
dx(2x + 3)2.
ddx
(2x + 3)2 = limh→0
(2(x + h) + 3
)2 − (2x + 3)2
h
= limh→0
(2x + 2h + 3 + 2x + 3)(2x + 2h + 3− 2x− 3)h
(see * below)
= limh→0
(4x + 2h + 6)(2h)h
= limh→0
(8x + 4h + 12)
= 8x + 12
Solution should be input 8x+12 or equivalent
* Confused by that step? Recall the difference of squares: a2 − b2 = (a + b)(a− b).
Let a =(2(x + h) + 3
)and b = (2x + 3). Then
a2 − b2 =(2(x + h) + 3
)2 − (2x + 3)2 =
((2(x + h) + 3
)+ (2x + 3)
)((2(x + h) + 3
)− (2x + 3)
)= (2x + 2h + 3 + 2x + 3)(2x + 2h + 3− 2x− 3)
= (4x + 2h + 6)(2h)
Of course, if you prefer, you can expand the terms:(2(x + h) + 3
)2 − (2x + 3)2 = (2x + 2h + 3)2 − (2x + 3)2
= 4x2 + 4xh + 6x + 4hx + 4h2 + 6h + 6x + 6h + 9− 4x2 − 12x− 9
= 8hx + 12h + 4h2
= (4x + 2h + 6)(2h)
Use whichever method you find easiest.
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I Question 8. Find the equation of the tangent line with slope 4
Find the coordinates of the point on the curve with tangent slope of 4:
8x + 12 = 4⇒ x1 = −1⇒ y1 =(2(−1) + 3
)2= 1
Then the equation of the line is:
y− y1x− x1
= 4
⇒ y− 1x− (−1)
= 4
⇒ y = 4x + 5
Solution should be input 4x+5 or equivalent
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I Question 1. limx→−1
(3x + 2)3 − x2
This is a polynomial function and continuous everywhere.The limit may be found by evaluating the function at x = −1
limx→−1
(3x + 2)3 − x2 = (3(−1) + 2)3 − (−1)2
= −2
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I Question 2. Evaluate limh→2
√h2 − 4
The limit does not exist. limh→2+
√h2 − 4 = 0 but the left limit
does not exist since√
h2 − 4 is not defined for the region to theleft of 2 (there is no real solution to the square root of anegative number.)
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I Question 3. Evaluate lims→1/3
√1 + 9s2
1 + 3s
No discontinuities in the region of s = 1/3 so
lims→1/3
√1 + 9s2
1 + 3s=
√1 + 9
(13
)2
1 + 3(
13
)=
√2
2
=1√2
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I Question 4. limp→0
3p2 − 2pp
limp→0
3p2 − 2pp
= limp→0
3p− 2 = −2
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I Question 5. limy→−2
8 + y3
y + 2
limy→−2
8 + y3
y + 2= lim
y→−2
���(y + 2)(y2 − 2y + 4)
���y + 2
= limx→−2
y2 − 2y + 4
= (−2)2 − 2(−2) + 4
= 12
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I Question 6. limn→2
n2 − 3n + 2n2 + n− 6
limn→2
n2 − 3n + 2n2 + n− 6
= limn→2
����(n− 2)(n− 1)����(n− 2)(n + 3)
=2− 12 + 3
= 15
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I Question 7. f (x) = 3x2 + 2x. Determine f ′(x).
f ′(x) = limh→0
f (x + h)− f (x)h
= limh→0
3(x + h)2 + 2(x + h)−(3x2 + 2x
)h
= limh→0
��3x2 + 6xh + 3h2 +��2x + 2h−��3x2 −��2xh
= limh→0
6x�h + 3h�2 + 2�h
�h= 6x + 2
Solution should be input 6x+2 or equivalent
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I Question 8. s(t) =√
t + 7. s′(t). Determine s′(t).
s′(t) = limh→0
s(t + h)− s(t)h
= limh→0
√t + h + 7−
√t + 7
h
= limh→0
√t + h + 7−
√t + 7
h·√
t + h + 7 +√
t + 7√t + h + 7 +
√t + 7
= limh→0
t + h + 7− t− 7
h(√
t + h + 7 +√
t + 7)
= limh→0
1√t + h + 7 +
√t + 7
=1
2√
t + 7
Solution should be input 1/2/(t+7)^(1/2) or 0.5/sqrt(t+7) or equivalent
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I Question 1. limx→−1
(2x + 3)2 − x3
This is a polynomial function and continuous everywhere.The limit may be found by evaluating the function at x = −1
limx→−1
(2x + 3)2 − x3 = (2(−1) + 3)2 − (−1)3
= 2
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I Question 2. Evaluate limt→3
√t2 + 9
The function is continuous everywhere in the region surroundingt = 3.The limit may be found by evaluating the function at t = 3
limt→3
√t2 + 9 =
√(3)2 + 9 =
√18 = 3
√2
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I Question 3. Evaluate lims→2
√4− s2
2− s
The limit does not exist.
lims→2+
√4− s2
2− sis not defined for the region to the left of 2 (there
is no real solution to the square root of a negative number.)
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I Question 4. limq→0
q + 5q2
5q
limq→0
q + 5q2
5q= lim
q→0
1 + 5q5
=15
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I Question 5.
limw→1
w3 − 1w− 1
= limw→1
����(w− 1)(
x2 + w + 1)
���w− 1
= limw→1
(x2 + w + 1
)= 3
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I Question 6.
The limit does not exist.The numerator approaches −6 as m→ 1. The denominator approaches 0 as m→ 1 but m2 − 3m + 2approaches 0 through positive values as m→ 1− and through negative values as m→ 1+. Thus
limm→1−
m2 − 3m− 4m2 − 3m + 2
=→ −6→ 0+
= −∞
limm→1+
m2 − 3m− 4m2 − 3m + 2
=→ −6→ 0−
= ∞
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I Question 7. s(t) = 2t2 + 3t. Determine s′(t).
s′(t) = limh→0
s(t + h)− s(t)h
= limh→0
2(t + h)2 + 3(t + h)− 2t2 − 3th
= limh→0
��2t2 + 4th + 2h2 +�3t + 3h−��2t2 −�3th
= limh→0
4t + 2h + 3
= 4t + 3
Solution should be input 4t+3 or equivalent
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I Question 8. f (x) = f (x) =√
x− 5. Determine f ′(x).
f ′(x) = limh→0
f (x + h)− f (x)h
= limh→0
√x + h− 5−
√x− 5
h
= limh→0
√x + h− 5−
√x− 5
h·√
x + h− 5 +√
x− 5√x + h− 5 +
√x− 5
= limh→0
x + h− 5− x + 5
h(√
x + h− 5 +√
x− 5)
= limh→0
1√x + h− 5 +
√x− 5
=1
2√
x− 5
Solution should be input 1/(x-5)^0.5 or equivalent
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I Question 1. Evaluate limx→2
x2 + 3x− 4
The function f (x) = x2 + 3x− 4 is a polynomial and continuous everywhere.
The limit is found by evaluating the function at x = 2:
limx→2
x2 + 3x− 4 = (2)2 + 3(2)− 4
= 6
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I Question 2. Evaluate limh→−4
√h2 + 9
The function f (x) = limh→−4
√h2 + 9 is continuous everywhere.
(For all real h, h2 + 9 is positive so the square root exists.)
The limit is found by evaluating the function at x = −4:
limh→−4
√h2 + 9 =
√(−4)2 + 9
=√
16 + 9
= 5
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I Question 3. Evaluate limx→2
√x2 − 4
As x approaches 2 from the left (x → 2 +−), x2 < 4 and the sign of x2 − 4 is negative.
Thus,√
x2 − 4 is not defined for values of x that are close to, but less than, 2.
Then, limx→2−
√x2 − 4 does not exist.
It follows that limx→2
√x2 − 4 does not exist either
(The right limit does exist and is equal to 0.)
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I Question 4. Evaluate limp→3
p2 − 9p− 3
At p = 3, both the numerator and the denominator of the fraction are equal to 0.
This is a limit of the type00
so we need to perform some algebra.
Factoring expressions of this type is generally quite easy:
Since p2 − 9 = 0 for p = 3, it follows that p− 3 is a factor of p2 − 9.
Then
limp→3
p2 − 9p− 3
= limp→3
(p− 3)(p + 3)p− 3
= limp→3
p + 3
= 6
Note that although p→ 3, p is never actually equal to 3 so p− 3 remains non-zeroand we can divide numerator and denominator by p− 3.
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I Question 5. Evaluate limy→−4
y2 + y− 122y2 + 7y− 4
At y = −4, both the numerator and the denominator of the fraction are equal to 0.
This is a limit of the type00
so we should factor the numerator and the denominator
and cancel any (non-zero) common terms.
Since y2 + y− 12 = 0 when y = −4, it follows that y− (−4) = y + 4 is a factor ofy2 + y− 12.
Similarly, y + 4 is a factor of 2y2 + 7y− 4.
Then
limy→−4
y2 + y− 122y2 + 7y− 4
= limy→−4
���(y + 4)(y− 3)���(y + 4)(2y− 1)
= limy→−4
(y− 3)(2y− 1)
(continuous at y = −4)
=−7−9
=79
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I Question 6. Evaluate limy→−4
y2 + y− 122y2 + 7y− 4
At y = −4, both the numerator and the denominator of the fraction are equal to 0.
This is a limit of the type00
so we should factor the numerator and the denominator
and cancel any (non-zero) common terms.
Since y2 + y− 12 = 0 when y = −4, it follows that y− (−4) = y + 4 is a factor ofy2 + y− 12.
Similarly, y + 4 is a factor of 2y2 + 7y− 4.
Then
limy→−4
y2 + y− 122y2 + 7y− 4
= limy→−4
���(y + 4)(y− 3)���(y + 4)(2y− 1)
= limy→−4
(y− 3)(2y− 1)
(continuous at y = −4)
=−7−9
=79
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I Question 7. Find the derivative of s(t) = 3t− 5t2 using the definition of the derivative:
s′(t) = limh→0
s(t + h)− s(t)h
= limh→0
3(t + h)− 5(t + h)2 −[3t− 5t2]
h
= limh→0
�3t + 3h−��5t2 − 10th− 5h2 −�3t +��5t2
h
= limh→0
3− 10t− 5h
= 3− 10t
Solution should be input 3-10t or equivalent
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I Question 8. Find the derivative of f (x) =√
x− 2 using the definition of the derivative:
f ′(x) = limh→0
f (x + h)− f (x)h
= limh→0
√(x + h)− 2−
√x− 2
h
= limh→0
√(x + h)− 2−
√x− 2
h·√(x + h)− 2 +
√x− 2√
(x + h)− 2 +√
x− 2
= limh→0
1h· (x + h− 2)− (x− 2)√
x + h− 2 +√
x− 2
= limh→0
1h· h√
x + h− 2 +√
x− 2
= limh→0
1√x + h− 2 +
√x− 2
=1
2√
x− 2
Solution should be input 2/(x-2)^0.5 or equivalent
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I Question 1. Find the line tangent to y = 3x2 + 4x + 2 at (−1, 1).
dydx
= 6x + 4
dydx
∣∣∣∣x=−1
= 6(−1) + 4 = −2
The line with slope −2 through (−1, 1) is given by:
y− y1x− x1
= m
⇒ y− 1x + 1
= −2
⇒ y− 1 = −2(x + 1)
⇒ y = −2x− 1
Solution should be input -2x-1 or equivalent.
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I Question 2. Find the line normal to xy + y2 + 2 = 0 at (3,−1)
xy + y2 + 2 = 0
⇒ xy′ + y + 2yy′ = 0
⇒ y′ = − y2y + x
⇒ y′∣∣(3,−1) = −
−12− 3
= 1
The slope of the line normal to the curve at (3,−1) is −1, so its equation is:
y− 1x + 3
= −1
⇒ y = −(x + 3) + 1
⇒ y = −x− 2
Solution should be input -x-2 or equivalent.
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I Question 3. Find the line normal to s(t) =√
8t + 1 with a slope of −3/4
First, find the location on the curve with slope 4/3:
s(t) = (8t + 1)1/2
s′(t) =12(8t + 1)−1/2 · 8 =
43
4√8t + 1
= −43
√8t + 1 = 3
8t + 1 = 9
t = 1
The curve has a slope of 4/3 when t = 1.
s(1) =√
8(1) + 1 = 3 so the normal to the curve has a slope of −3/4 at the point
(1, 3) Find the equation of the line:
s− 3t− 1
=34
s =34(t− 1) + 3
Solution should be input 27x/4+2 or equivalent.
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I Question 4. Find the line tangent to y =1
(x + 1)2 with a slope of 27/4
y = (x + 1)−2
dydx
= −2(x + 1)−3 =−2
(x + 1)3
Now, find the point on the curve where the slope is 27/4:
−2(x + 1)3 =
274
⇒ (x + 1)3 = − 827
⇒ x + 1 =3
√− 8
27= −2
3
⇒ x = −53
y(−5
3
)=
1(− 5
3 + 1)2 =
94
The curve has a slope of 27/4 at the point (5/3, 9/4).Find the equation of the line:
y− 94
x + 53=
274
⇒ y =274
(x +
53
)+
94
⇒ y =274
x +274× 5
3+
94
⇒ y =274
x +272
⇒ y =274(x + 2)
Solution should be input 27x/4+2 or equivalent.
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I Question 1. 5y− x2 = 5,dxdt
and x = 2. Finddydt
∣∣∣∣x=2
.
Differentiate both x and y with respect to t
5y− x2 = 5
5dydt− 2x
dxdt
= 0
dydt
=25· x · dx
dt
=25(2)(3)
=125
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I Question 2.
Let x be the distance from the bottom of the ladder to the bottom of the wall and lety be the distance from the top of the ladder to the bottom of the wall. Both x and yare functions of time, t. Then, using the Pythagorean Theorem, x and y are relatedby the formula
x2 + y2 = 102
When x = 6, y = 8 since 62 + 82 = 102. Also,dxdt
= 1 is given.
Differentiating with respect to t gives:
2xdxdt
+ 2ydydt
= 0
dydt
= − xy· dx
dt
= −68(1)
= −34
The distance y from the top of the ladder to the bottom of the wall is decreasing at a
rate of34
m/s.
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I Question 3.
V and r are both functions of time, t, related by V =43
πr3. Differentiating with
respect to t gives:
dVdt
=4π
33r2 dr
dtdrdt
=dVdt
14πr2
= 0.09 · 14π(0.6)2
= 0.019894 m/s
The radius is increasing at a rate of 19.894 mm/s so the diameter is increasing attwice that rate: 39.789 mm/s.
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I Question 1. Use the product rule to differentiate y = (3− x)(
3 +1x
)
(uv)′ = uv′ + vu′. Let u = 3− x and v =
(3 +
1x
)
y′ = (3− x)(−x−2
)+
(3 +
1x
)(−1)
= − 3x2�
��+1x− 3
���− 1x
= −3(
1 +1x2
)
Solution should be input -3(1+1/x^2) or equivalent.
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I Question 2. Use the quotient rule to differentiate s(t) =2t2 + 1t2 + 1
.
(uv
)′=
vu′ − vu′
v2 . Let u = 2t2 + 1 and v = t2 + 1
s′ =(t2 + 1
) (2t2 + 1
)′ − (2t2 + 1) (
t2 + 1)′
(t2 + 1)2
=
(t2 + 1
)(4t)−
(2t2 + 1
)(2t)
(t2 + 1)2
=��4t3 + 4t−��4t3 − 2t
(t2 + 1)2
=2t
(t2 + 1)2
Solution should be input 2t/(t^2+1)^2 or equivalent.
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I Question 3. Differentiate g(x) =√
3− x2.
g′(x) =12
(3− x2
)−1/2· (−2x)
=−x√3− x2
Solution should be input -x/(3-x^2)^(1/2) or even -x/sqrt(3-x^2) or equivalent.
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I Question 4. Differentiate x3 + 3y− y2 = 17 with respect to x
3x2 + 3y′ − 2yy′ = 0
y′(3− 2y) = −3x2
dydx
=3x2
2y− 3
Solution should be input 3x^2/(2y-3) or equivalent.
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I Question 4. Evaluate the slope of the curve at (3, 5)
dydx
∣∣∣∣(3,5)
=3(3)2
2(5)− 1=
277
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I Question 5. Determine the line normal to the curve
y = x2 − 3x− 2 at the point (4, 2)
y′ = 2x− 3
y′∣∣x=4 = 8− 3 = 5
y− 2x− 4
= −15
y− 2 =4− x
5
y =14− x
5
Solution should be input (14-x)/5 or equivalent.
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I Question 6. Find the maximum of p(x) = 23x− 0.1x2
p′(x) = 23− 0.2x = 0
x = 115
p(115) = 23(115)− 0.1(115)2
= 1322.5
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I Question 7. (The volume of a sphere is V = 43 πr3, 1 m = 1000 mm.)
This is a related rates problem; both V and r are functions of time, t.
V = 43 πr3
dVdt
=43
π · 3r2 drdt
= 4πr2 drdt
= 4π(1.8)2(0.028)
= 1.14
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I Question 1. Use the product rule to differentiate y =(
3− x2)(
3 +1x2
)
(uv)′ = uv′ + vu′. Let u = 3− x2 and v =
(3 +
1x2
)
y′ = (3− x2)(−2x−3
)+
(3 +
1x2
)(−2x)
= − 6x3�
��+2x− 6x
���− 2x
= −6(
x +1x3
)
Solution should be input -6*(x+1/x^3) or equivalent.
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I Question 2. Use the quotient rule to differentiate s(t) =2t2 + 1t2 + 1
.
(uv
)′=
vu′ − vu′
v2 . Let u = 3− t2 and v = 3 + t2
s′ =(3 + t2) (3− t2)′ − (3− t2) (3 + t2)′
(3 + t2)2
=
(3 + t2) (−2t)−
(3− t2) (2t)
(3 + t2)2
=−6t���−2t3 − 6t���+2t3
(3 + t2)2
=−12t
(3 + t2)2
Solution should be input -12t/(3+t^2)^2 or equivalent.
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I Question 3. Differentiate g(x) =√
1 + 3x2.
g′(x) =12
(1 + 3x2
)−1/2· (6x)
=3x√
1 + 3x2
Solution should be input 3x/sqrt(1+3x^2) or even -x/sqrt(3-x^2) or equivalent.
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I Question 4. Differentiate x2 − y− y2 + 2 = 0 with respect to x
2x +dydx− 2y
dydx
= 0
dydx
(1− 2y) = −2x
dydx
=2x
2y− 1
Solution should be input 2x/(2y-1) or equivalent.
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I Question 4. Evaluate the slope of the curve at (2, 3)
dydx
∣∣∣∣(2,3)
=2(2)
2(3)− 1=
45
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I Question 5. Evaluate lims→−1
1 + 3s + s2
s
y′ = 6x− 4
y′∣∣x=2 = 12− 4 = 8
y− 1x− 2
= −18
y− 1 =2− x
8
y =10− x
8
Solution should be input (10-x)/8 or equivalent.
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I Question 6. Find the maximum of p(x) = 23x− 0.1x2
p′(x) = 23− 0.2x = 0
x = 115
p(115) = 23(115)− 0.1(115)2
= 1322.5
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I Question 7. (The volume of a sphere is V = 43 πr3, 1 m = 1000 mm.)
This is a related rates problem; both V and r are functions of time, t.
V = 43 πr3
dVdt
=43
π · 3r2 drdt
= 4πr2 drdt
= 4π(3.7)2(0.046)
= 7.914
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I Question 1.
∫4t3 − 3t2 + 2t− 1 dt = t4 − t3 + t2 − t + C
Solution should be input t^4-t^3+t^2-t or equivalent
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I Question 2.
∫3 f (x) dx = 3
∫f (x) dx == 3
∫x2 − 2x dx = 3
(x3
3− x2
)+ C = x3 − 3x2 + C
Alternatively,∫3 f (x) dx =
∫3(
x2 − 2x)
dx =∫
3x2 − 6x dx = x3 − 3x2 + C
Solution should be input x^3-3x^2 or equivalent
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I Question 3.
∫ 1x3 −
1x2 dx =
∫x−3 − x−2 dx
=x−2
−2− x−1
−1+ C
= − 12x2 +
1x+ C
Solution should be input -1/(2x^2)+1/x or equivalent
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I Question 4.
∫ √x3 +
3√
x2 dx =∫
x3/2 + x2/3 dx
=x5/2
5/2− x5/3
5/3+ C
= −2x5/2 + 3x5/3
5+ C
Solution should be input (2x^(5/2)+3x^(5/3))/5 or equivalent
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I Question 5.
∫x(√
x +1√x
)dx =
∫x3/2 + x1/2 dx
=x5/2
5/2+
x3/2
3/2+ C
=2x5/2
5+
2x3/2
3+ C
= 2x√
x(
x5+
13
)+ C
Solution should be input 2x^(5/2)/5+2x^(3/2)/3 or equivalent
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I Question 6.
∫(t + 1)2 dt =
∫t2 + 2t + 1 dt
=t3
3+ t2 + t + C
Solution should be input t^3/3+t^2+t or equivalent
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I Question 7.
Let u = s + 1
Then du = ds
∫ ds(s + 1)2 =
∫ duu2
=∫
u−2 du
=u−1
−1+ C
= − 1u+ C
= − 1s + 1
+ C
Solution should be input -1/(s+1) or equivalent
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I Question 8.
∫3√
p2 − 2p + 1 dp =∫
3√(p− 1)2 dp
=∫
(p− 1)2/3 dp
Let u = p− 1
Then du = dp
∫(p− 1)2/3 dp =
∫u2/3 du
=u5/3
5/3+ C
=3u5/3
5+ C
=3(p− 1)5/3
5+ C
Solution should be input 3(p-1)^(5/3)/5 or equivalent
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I Question 9.
Let u = 1− x3
Thendudx
= −3x2
and dx =du−3x2
∫x2√
1− x3 dx =∫
x2√
1− x3 du−3x2
= −13
∫u1/2 du
= −13
[u3/2
3/2
]+ C
= −2(1− x3)3/2
9+ C
Solution should be input -2(1-x^3)^(3/2)/9 or equivalent
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I Question 10.
The curve with slope 3√
x is y =∫
3√
x dx =∫
x1/3 dx =3x4/3
4+ C.
The curve passes through (8, 14) so y(8) = 14.Then,
y(8) =3(8)4/3
4+ C =
3(2)4
4+ C =
3(16)4
+ C = 12 + C = 14⇒ C = 2
The equation is given by
y =3x4/3
4+ 2
Solution should be input 3x^(4/3)/4+2 or equivalent
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I Question 1.
∫ 3
1t3 dt =
[x4
4
]3
1
=34
4− 14
4= 20
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I Question 2.
Let u = 1 + xThen du = dx
∫ 3
0
dx√1 + x
=∫ x=3
x=0
du√u
=∫ x=3
x=0u−1/2 du
=
[u1/2
1/2
]x=3
x=0
= 2[(1 + x)1/2
]3
0
= 2[41/2 − 11/2
]= 2
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I Question 3.
∫ 1
−2s(
1− s2)
ds =∫ 1
−2s− s3 ds
=
[s2
2− s4
4
]1
−2
=
[(12− 1
4
)−(
42− 16
4
)]=
14− (−2)
=94
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I Question 4.
∫ 3
−2x2 +
1x2 dx =
∫ 3
−2x2 + x−2 dx
=
[x3
3+
x−1
−1
]3
−2
=
[x3
3− 1
x
]3
−2
=
(33
3− 1
3
)−((−2)3
3− 1−2
)=
273− 1
3+
83− 1
2
=343− 1
2
=656
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I Question 5.
To solve this question, you need to recognise that (2y− 1)2 = 4y2 − 4y + 1. Then,
Let u = 2y− 1
Thendudy
= 2
and dy =du2
∫ 14
1
dy3√
4y2 − 4y + 1=∫ 14
1
dy3√(2y− 1)2
=∫ y=14
y=1
13√
u2· du
2
=12
∫ y=14
y=1u−2/3 du
=12
[u1/3
1/3
]y=14
y=1
=32
[(2y− 1)1/3
]14
1
=32
[271/3 − 11/3
]= 3
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I Question 6.
Let u = x2 − 1
Thendudx
= 2x
and dx =du2x
∫ 3
0
x3√
x2 − 1dx =
∫ x=3
x=0
x3√
udu2x
=12
∫ x=3
x=0u−1/3 du
=12
[u2/3
2/3
]x=3
x=0
=34
[(x2 − 1
)2/3]3
0
=34
[82/3 − (−1)2/3
]=
34[4− 1]
=94
Note that (−1)1/3 = −1⇒ (−1)2/3 = (−1)2 = 1
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I Question 7.
Let u = x2 + 9
Thendudx
= 2x
and dx =du2x
∫ 4
0x√
x2 + 9 dx =∫ x=4
x=0x√
udu2x
=12
∫ x=4
x=0u1/2 du
=12
[u3/2
3/2
]x=4
x=0
=13
[(x2 + 9
)3/2]4
0
=13
[253/2 − 93/2
]=
13[125− 27]
=983
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I Question 8.
Let u = 2x2 − 3x + 1
Thendudx
= 4x− 3
and dx =du
4x− 3
∫ 5
0(6− 8x)
√2x2 − 3x + 1 dx =
∫ x=5
x=0(6− 8x)
√u
du4x− 3
= −2∫ x=5
x=0u1/2 du
= −2
[u3/2
3/2
]x=5
x=0
= −43
[(2x2 − 3x + 1
)3/2]5
0
= −43
[(2(5)2 − 3(5) + 1
)3/2−(
2(0)2 − 3(0) + 1)3/2
]= −4
3
[363/2 − 13/2
]= −4
3[216− 1]
= −8603
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I Question 1.
∫ 4t4 − 3t3
tdt =
∫4t3 − 3t2 dt
= t4 − t3 + C
Solution should be input 4t^4-3t^3 or equivalent.
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I Question 2.
y =∫
f (x) dx
=∫
2x3 − 3x dx
=x4
2− 3x2
2+ C
y(2) = 9
⇒ 8− 6 + C = 9
⇒ C = 7
y =x4
2− 3x2
2+ 7
Solution should be input x^4/2-3x^2/2+7 or equivalent.
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I Question 3.
∫ (y5/3 +
1y3/5
)dy =
∫ (y5/3 + y−3/5
)dy
=3y8/3
8+
5y2/5
2+ C
Solution should be input 3y^(8/3)/3+5y^(2/5)/2 + C or equivalent.
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I Question 4.
∫ 1
−27− 2x + x3 dx =
[7x− x2 +
x4
4
]1
−2
=
(7− 1 +
14
)−(−14− 4 +
164
)= 20.25
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I Question 5.
Let u = 2t− 1
Thendudt
= 2
and dt =du2
∫ 14
0
dt3√
4t2 − 4t + 1=∫ t=14
t=0
du2u2/3
=12
∫ t=14
t=0u−2/3 du
=12
[u1/3
1/3
]t=14
t=0
=32
[(2t− 1)1/3
]14
0
=32
[(27)1/3−(−1)1/3
]=
32[3− (−1)]
= 6
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I Question 6.
Let u = s3 + 3s2 − 4
Thenduds
= 3s2 + 6s
and ds =du
3s2 + 6s
∫ 2
−2(s2 + 2s)
√s3 + 3s2 − 4 ds =
∫ s=2
s=−2(s2 + 2s)
√u
du3s2 + 6s
=13
∫ s=2
s=−2u1/2 du
=13
[u3/2
3/2
]s=2
s=−2
=29
[(s3 + 3s2 − 4
)3/2]2
−2
=29
[(8 + 12− 4)3/2 − (−8 + 12− 4)3/2
]=
29
[(16)3/2 − (0)3/2
]=
29[64]
=128
9
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I Question 1.
∫ 4s4 + 2s2
sds =
∫4s3 + 2s ds
= s4 + s2 + C
Solution should be input s^4+s^2 or equivalent.
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I Question 2.
y =∫
h(s)) ds
=∫
s3 + s ds
=s4
4+
s2
2+ C
y(−1) = 0
⇒ 14+
12+ C = 0
⇒ C = −34
y =s4
4+
s2
2− 3
4
Solution should be input s^4/4+s^2/2-3/4 or equivalent.
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I Question 3.
∫ (y4/5 − 1
y5/4
)dy =
∫ (y4/5 − y−5/4
)dy
=y9/5
9/5− y−1/4
−1/4+ C
=5y9/5
9+ 4y−1/4 + C
or =5y9/5
9+
4y1/4 + C
Solution should be input 5y^(9/5)/9+4y^(-1/4) + C or equivalent.
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I Question 4.
∫ 2
−1t− 2t3 + 3 dt =
[t2
2− t4
2+ 3t
]2
−1
=
(42− 16
2+ 6)−(
12− 1
2− 3)
= 3
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I Question 5.
Let u = s− 1
Thenduds
= 1
and ds = du
∫ 9
0
ds3√
s2 − 2s + 1=∫ s=9
s=0
du3√
u2
=∫ s=9
s=0u−2/3 du
=
[u1/3
1/3
]s=9
s=0
= 3[(s− 1)1/3
]9
0
= 3[(8)1/3−(−1)1/3
]= 3 [2− (−1)]
= 9
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I Question 6.
Let u = 2x2 − 3x + 1
Thendudx
= 4x− 3
and dx =du
4x− 3
∫ 5
0(6− 8x)
√2x2 − 3x + 1 dx =
∫ x=5
x=0(6− 8x)
√u
dx4x− 3
= −2∫ x=5
x=0u1/2 du
= −2
[u3/2
3/2
]x=5
x=0
= −43
[(2x2 − 3x + 1
)3/2]5
0
= −43
[(50− 15 + 1)3/2 − (1)3/2
]= −4
3[216− 1]
= −8603
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I Question 1.
∫ 4t4 + 3t3
t3 dt =∫
4t + 3 dt
= 2t2 + 3t + C
Solution should be input 2t^2+3t+C or equivalent.
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I Question 2.
y =∫
g(t) dt
=∫
t2 − 4t dt
=t3
3− 2t2 + C
y(3) = −6
⇒ 33
3− 2(3)2 + C = −6
9− 18 + C = −6
⇒ C = 3
y =t3
3− 2t2 + 3
Solution should be input t^3/3-2t^2+3 or equivalent.
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I Question 3.
∫ (y3/4 − 1
y4/3
)dy =
∫ (y3/4 − y−4/3
)dy
=y7/4
7/4− y−1/3
−1/3+ C
=4y7/4
7+ 3y−1/3 + C
or =4y7/4
7+
3y1/3 + C
Solution should be input 4y^(7/4)/7+3y^(-1/3) + C or equivalent.
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I Question 4.
Let u = s3 + 3s2 − 4
Thenduds
= 3s2 + 6s
and ds =du
3s2 + 6s
∫ 2
−2(s2 + 2s)
√s3 + 3s2 − 4 ds =
∫ s=2
s=−2(s2 + 2s)
√u
du3s2 + 6s
=13
∫ s=2
s=−2u1/2 du
=13
[u3/2
3/2
]s=2
s=−2
=29
[(s3 + 3s2 − 4
)3/2]2
−2
=29
[(8 + 12− 4)3/2 − (−8 + 12− 4)3/2
]=
29
[(16)3/2 − (0)3/2
]=
29[64]
=128
9
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I Question 5.
Let u = x + 2
Then dx = du
∫ 6
−3
dx3√
x2 + 4x + 4=∫ 6
−3
dx3√(x + 2)2
=∫ 6
−3
dx(x + 2)2/3
=∫ x=6
x=−3u−2/3 du
=
[u1/3
1/3
]x=6
x=−3
= 3[(x + 2)1/3
]−
36
= 3[81/3 − (−1)1/3
]= 3 [2− (−1)]
= 9
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I Question 6.
Let u = 3t + t3
Thendudt
= 3 + 3t2
and dt =du
3(t2 + 1)
∫ 1
0(t2 + 1)
√3t + t3 dt =
∫ x=1
x=0(t2 + 1)
√u
dt3(t2 + 1)
=13
∫ x=1
x=0u1/2 du
=13
[u3/2
3/2
]x=1
x=0
= −29
[(3t + t3
)3/2]1
0
= −29
[(4)3/2 − (0)3/2
]=
169
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I Question 1.
Find the area bounded by y = x2 − 2x− 3 and the line y = 0.Note that y = x2 − 2x− 3 is a parabola that opens upwards (the x2 coefficient is positive)so the area enclosed by the parabola and the line y = 0 will be below the x-axis.
Find the intersects of the curve y = x2 − 2x− 3 with the line y = 0:
x2 − 2x− 3 = 0
⇒ (x− 3)(x + 1) = 0
Thus, the intersects are at x = −1 and x = 3. The area looks like this:
yy = x2 − 2x− 3
y = 0
(−1, 0) (3, 0)
The required area is:
A =∫ 3
−10−
(x2 − 2x− 3
)dx
=∫ 3
−1−x2 + 2x + 3 dx
=
[− x3
3+ x2 + 3
]3
−1
=
[−27
3+ 9 + 9
]−[
13+ 1− 3
]= 9−
[−1
23
]= 10 2
3
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I Question 2.
Find the area bounded by y = x2 and the line y = x.Note that y = x2 is a parabola, through the origin, that opens upwards (the x2 coefficient ispositive).
Find the intersects of the curvey = x2 with the line y = x:
x2 = x
⇒ x2 − x = 0
⇒ x(x− 1) = 0
Thus, the intersects areat x = 0 and x = 1. x
y y = x2
y = x
(0, 0)
(1, 1)
The required area is:
A =∫ 1
0x− x2 dx
=
[x2
2− x3
3
]1
0
=
[12− 1
3
]− [0− 0]
=16
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I Question 3.
Find the area bounded by y = (x− 1)2 − 2 and y = 3− x2.
y = (x − 1)2 − 2 is a parabola that opens upwards since the x2 coefficient is positive.(x− 1) indicates that the parabola is shifted to the right by 1 unit and the −2 indicatesthat the parabola is shifted down by 2 units.
y = −x2 + 3 is a parabola that opens downwards since the x2 coefficient is negative.The +3 indicates that the parabola is shifted up by 3 units.
Find the intersections of the twoparabolas:
(x− 1)2 − 2 = −x2 + 3
⇒ x2 − 2x + 1− 2 = −x2 + 3
⇒ 2x2 − 2x− 4 = 0
⇒ 2(x + 1)(x− 2) = 0
Thus, the parabolas intersect atx = −1 and x = 2.
x
y
y = (x− 1)2 − 2
y = −x2 + 3
(−1, 2)
(2,−1)
The required area is:
A =∫ 2
−1
[−x2 + 3
]−[(x− 1)2 − 2
]dx
=∫ 2
−1−2x2 + 2x + 4 dx
= −2∫ 2
−1x2 − x− 2 dx
= −2[
x3
3− x2
2− 2x
]2
−1
= −2[
83− 4
2− 4]− 2
[−1
3− 1
2+ 2]
= −2[
93− 3
2− 6]
= 9
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I Question 4.
Find the area bounded by f (x) = x3 − 3x2 − x + 3 and the x-axis between x = −1 and x = 3.
f (x) is a cubic polynomial with a positive leading coefficient so f (x) → −∞ as x → −∞ (andf (x)→ ∞ as x → ∞).
As x → −1− (as x approaches −1 from the left), f (x) is negative. f (x) crosses the x-axisat x = −1 and remains positive until it crosses the x-axis again at x = 1. Thus, f (x) > 0 for−1 < x < 1. f (x) remains negative until it crosses the x-axis into positive territory again at x = 2.
Thus,
f (x)
< 0 if x < −1= 0 if x = −1> 0 if − 1 < x < 1= 0 if x = 1< 0 if 1 < x < 2= 0 if x = 2> 0 if x > 2
x
f (x)
f (x) = x3 − 3x2 − x + 3
(−1, 0)
(2, 0)
The required area is:
A = A1 + A2
=∫ 1
−1
(x3 − 3x2 − x + 3
)− 0 dx +
∫ 2
10−
(x3 − 3x2 − x + 3
)dx
=
[x4
4− x3 − x2
2+ 3x
]1
−1
+
[− x4
4+ x3 +
x2
2− 3x
]2
1
=
[(14− 1− 1
2+ 3)−(
14+ 1− 1
2− 3)]−[(−16
4+ 8 +
42− 6)−(−1
4+ 1 +
12− 3)]
= 5 34
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I Question 5.
Find the area bounded by y = x2, y = − x2
4+ 5 and y = 2x.
y = x2 opens upwards and y = − x2
4+ 5 opens downwards. Find their points of intersection:
x2 = − x2
4+ 5
5x2
4= 5
x2 = 4
Thus, x = ±2 and the parabolas intersectat (−2, 4) and (2, 4).The points of intersection of y = x2 andy = 2x are:
x2 = 2x
x2 − 2x = 0
x(x− 2) = 0
x = 0 or 2
x
y
y = − x2
4+ 5
y = 2x
(0, 0)
(−2, 4) (2, 4)
y = x2
y = x2 and y = 2x intersect at (0, 0) and (2, 4).
Then the total required area is:
A = A1 + A2 =∫ 0
−2
[(− x2
4+ 5)− x2
]dx +
∫ 2
0
[(− x2
4+ 5)− 2x
]dx
= 5∫ 0
−21− x2
4dx−
∫ 2
0
x2
4+ 2x− 5 dx
= 5[
x− x3
12
]0
−2−[
x3
12+ x2 − 5x
]2
0
= −5[−2 +
812
]−[
812
+ 4− 10]
= 12
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I Question 1.
Let x be the amount of cable that has been wound upThe weight of cable between the elevator and the cable drum is 192(37.5− x) NThe force at the cable drum equals the weight of the cable and the weight of the elevator:F = 192(37.5− x) + 9000The work done in raising the elevator 18.0 m from the basement to the fifth floor is:∫ 18
0192(37.5− x) + 9000 dx =
∫ 18
016200− 192x dx
=[16200x− 96x2
]18
0
= 260496 N ·m= 260.5 kN ·m
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I Question 1.
The force at the cable drum equals the weight of the cable and the weight of the elevator:F = 192(37.5− x) + 7125 The work done in raising the elevator 15.0 from the fifth to thetenth floor is: ∫ 33
18192(37.5− x) + 7125 dx =
∫ 33
1814325− 192x dx
=[14325x− 96x2
]33
18
= 141435 N ·m= 141.44 kN ·m
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I Question 2.
First, find the spring constant:
f (x) = kx720 N = k(10m)
k = 72 N/m
The work is done stretching the bungee cord from its unstretched length of 50 m by 10 mto its final length of 60 m.
w =∫ b
af (x) dx
=∫ 10
072xdx
=[36x2
]10
0
= 3600 N ·m= 3.6 kN ·m
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