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Derivative Tests
Mathematics 100
Institute of Mathematics
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Outline
1 Increasing and Decreasing functions
2 Relative and Absolute Extrema
3 Concavity and Points of Inflection
4 Curve Sketching
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Intuitive Definition of Increasing and Decreasing Functions
On the interval(, 0),as the graph goes from left to right, thegraph is rising.
On the interval(0,+),as the graph goes from left to right, the
graph is falling.
The function isincreasingat (,0)anddecreasingat(0,+).
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Increasing and Decreasing Functions
Definition
A functionf isincreasingon an interval Iif for anyx1 andx2 in the
intervalI,
x1
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Illustration
x1 x2
y1
y2
x1< x
2
y1 < y2
x1 x2
y2
y1
x1< x
2
y1 > y2
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Application of the Derivative
Consider the tangent lines at the intervals where the function is
increasing:
Remark
When the graph is increasing on an interval I, then for any x I,theslope of the tangent line at that point is positive.
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Application of the Derivative
Consider the tangent lines at the intervals where the function is
decreasing:
Remark
When the graph is decreasing on an interval I, then for any x I,theslope of the tangent line at that point is negative.
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Application of the Derivative
Consider the tangent lines at the intervals where the function is
constant.
a b
Remark
At the interval(a,b), where the graph is neither increasing nordecreasing, the tangent line is the horizontal line (slope is zero).
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Theorem
Theorem
Letfbe differentiable on an interval(a,b).
1. Iff(x)> 0 for allx (a,b)thenfis increasing on(a, b).
2. Iff(x)< 0 for allx (a,b)thenfis decreasing on(a,b).
3. Iff(x) =0 for all x (a,b)thenfis constant on(a,b).
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Critical Numbers
Remark
For a continuous functionf,f(x)can change sign only atx -values
wheref(x) =0 or is undefined.
We call thesex-values thecritical numbersof f.
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Examples
1. Find the intervals on whichf(x) =x3
3
2x2
is increasing ordecreasing.
Solution:
We need to determine the intervals when f(x)is positive and negative.
To determine these intervals, find the critical number(s). (values of xwhenf(x) =0.
f(x) = 3x2 3x
0 = 3x
2
3x0 = 3x(x 1)
x = 0, 1
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Examples
Solution (Contd):Determine the sign off(x)(notf(x)) for the intervals(,0),(0, 1),and(1,+).
(,0) (0,1) (1,+)x + +
x 1 +
f(x) =3x(x 1) + +
Hence,fis increasing at the interval(,0) (1,+)anddecreasing at the interval(0, 1).
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Relative and Absolute Extrema
Definition
Letfbe a function defined at c.
1. f(c)is called arelative maximumof fif there exists an interval(a,b)containingcsuch that for allx (a,b)
f(x) f(c)
2. f(c)is called arelative minimumof fif there exists an interval(a,b)containingcsuch that for allx (a,b)
f(x) f(c)
3. f(c)is anabsolute maximumoff iff(x) f(c)for everyx domf.f(c)is anabsolute minimumof f iff(x) f(c)for everyx domf.
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Illustration
relative minimum
relative maximum
relative minimum
absolute minimumno absolute maximum
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FDT f R l i E
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FDT for Relative Extrema
First Derivative Test for Relative Extrema:
Letfbe continuous on the interval(a,b)in whichcis the only criticalnumber. Iffis differentiable on the interval (except possibly at c), then
1. Iff(x)> 0 forxc, thenf(c)
is a relative maximum.
2. Iff(x)< 0 forx 0 forx>c, thenf(c)is a relative minimum.
3. Iff(x)> 0 forx (a,b)such thatx =candf(x)
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Examples
1. Find all relative extrema of the function
f(x) =2x3 3x2 36x+14.
Solution:
Look for values of xwhere the sign of the first derivative changes.
Find the critical numbers: Set the derivative of f(x)equal to zero.
f(x) = 6x2 6x 36
= 6(x2 x 6)
= 6(x 3)(x+2) =0 x=3, 2
Sincef is defined for all real numbers, the only critical numbers of f arex= 2 andx=3.
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f ( ) 2 3 3 2 36 14
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f(x) =2x3 3x2 36x+14
Solution (Cont.):
(, 2) 2 (2,3) 3 (3,+)
f(x) 58 67
x+2 0 + + +x 3 0 +
f(x) =6(x+2)(x 3) + 0 0 +Using the First Derivative Test, f(2) =58 is a relative maximum andf(3) = 67 is a relative minimum.
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E l
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Examples
2. Find all relative extrema of the function f(x) =x4 4x3.
Solution:
f(x) =4x3 12x2 =4x2(x 3) =0 x=0,3
(,0) 0 (0,3) 3 (3,+)f(x) 0 27
x2 + 0 + + +
x 3 0 +
f
(x) =4x2
(x 3) 0 0 +Atx=0,f does not change sign from to+, vice versa. Hence theonly extremum isf(3) = 27 which is a relative minimum.
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SDT for Relative Extrema
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SDT for Relative Extrema
Second Derivative Test for Relative Extrema
Supposef exists on some open interval containing candf(c) =0.
1. Iff(c)> 0, thenf(c)is a relative minimum.2. Iff(c)< 0, thenf(c)is a relative maximum.
3. Iff(c) =0, then the test fails.
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Illustration
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Illustration
We use the second derivative test in the previous examples to
determine the relative extrema.
Example 1
Find the relative extrema off(x) =2x3 3x2 36x+14.
We need to find the values csuch thatf
(c) =0.(the criticalnumbers).x= 2 andx=3(Verify!)
Find the second derivative:
f(x) =12x 6
Evaluate the second derivative at the critical numbers:
f(2) =12(2) 6= 30< 0 relative maximumf(3) =12(3) 6= 30>0 relative minimum
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Illustration
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Illustration
Example 2
Find the relative extrema off(x) =x4 4x3.
We need to find the values csuch thatf(c) =0.(the criticalnumbers).x=0 andx=3(Verify!)
Find the second derivative:f(x) =12x2 24x
Evaluate the second derivative at the critical numbers:
f(3) =12(3)2 24(3) =36> 0 relative minimum
f
(0) =12(0) 24(0) =0 test failsApply the FDT forx=0.
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Definitions
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Definitions
Concavity
Letfbe differentiable on an open intervalI. The graph off is
1. concave upwardonI iff is increasing on the interval.
2. concave downwardonI iff is decreasing on the interval.
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Definitions
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Definitions
concave upward
f is increasingconcave downward
f is decreasing
1. A curve that lies above its tangent lines is concave upward.
2. A curve that lies below its tangent lines is concave downward.
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Definitions
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Definitions
Letfbe a function whose second derivative exists on an open interval I.
1. Iff(x)> 0 for allx I, then the graph of f is concave upward on I.
2. Iff(x)< 0 for allx I, then the graph of f is concave downward onI.
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Definitions
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Definitions
For a continuous functionfwe can find the intervals on which thegraph is concave upward and concave downward through these steps:
a) Find the second derivative.
b) Locate thexvalues at whichf(x) =0 or undefined.
c) Use thexvalues to determine the test intervals.d) Test the sign of f(x)in each of the test intervals.
Point of Inflection
The point wherein the concavity changes is called apoint of
inflection.
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Example
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Example
Determine the points of inflection and discuss the concavity of the
graph off(x) =x4 +x3 3x2 +1.
Solution:
Differentiating twice, we have
f(x) =12x2 +6x 6= 6(2x2 +x 1) =6(2x 1)(x+1)
By settingf(x) =0, we see the that the possible points of inflection
occur atx= 1 andx=1/2.
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Example
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Example
Solution (Cont.):
(, 1) 1
1, 12
1
2
12, +
f(x) 2 7
16
2x 1 0 +
x+1 0 + + +f(x) =6(2x 1)(x+1) + 0 0 +
Hence the graph is concave up on the interval(, 1)
1
2, +
and is
concave down on
1,1
2
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Curve Sketching (Examples)
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Curve Sketching (Examples)
1. Sketch the graph off(x) =x3 9x2 +15x 12.
Solution:
Solve the first and second derivatives:
f
(x) =3x2 18x+ 15= 3(x2 6x+ 5) =3(x 1)(x 5) =0 x=1, 5
f(x) =6x 18= 6(x 3) =0 x=3
f(1) = 5 andf(5) = 37 hence(1, 5)and (5, 37)are the criticalpoints of the graph of f.
f(3) = 21, so(3, 21)is a possible point of inflection.
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Curve Sketching (Examples)
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Curve Sketching (Examples)
Solution (Cont.):
(, 1) 1 (1, 3) 3 (3, 5) 5 (5,+)f(x) 5 21 37
x 1 0 + + + + +x 5 0 +
f(x) = 3(x1)(x5) + 0 0 +
f(x) = 6(x3) 0 + + +
graph of f(x) inc rel dec point of dec rel incconc down max conc down inflection conc up min conc up
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Curve Sketching (Examples)
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g ( p )
Solution (Cont.):
Graph:
3 2 1 1 2 3 4 5 6 7 8 9 10
40
35
30
25
20
15
10
5
5
10
0
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Curve Sketching (Examples)
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g ( p )
2. Sketch the graph off(x) = x3 +3x2 2.
Solution:
Derivatives:
f
(x) = 3x2 +6x=3x(x+2) =0 x=0, 2
f(x) = 6x+6= 6(x+1) =0 x=1
f(0) = 2 andf(2) =2, hence(0, 2)and (2, 2)are the critical points ofthe graph off.
Evaluating f atx=1, we get 0, so(1, 0)is a possible point of inflection.
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Curve Sketching (Examples)
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g ( p )
Solution (Cont.):
(,0) 0 (0, 1) 1 (1, 2) 2 (2,+)
f(x) 2 0 2
x 0 + + + + +
2 x + + + + + 0
f(x) =3x(2 x) 0 + + + 0
f(x) =6(1 x) + + + 0
graph of f(x) dec rel inc point of inc rel dec
conc up min conc up inflection conc down max conc down
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Curve Sketching (Examples)
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g ( p )
Solution (Cont.):
Graph:
1 1 2 3
2
1
1
2
3
0
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Lecture Exercises
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1 Determine the interval/s over which the graph of the function
f(x) =e2x 3ex +x 2 is increasing/decreasing.
2 Sketch the graphs of the following functions.
a. f(x) =x3 3x2 +4b. f(x) =x3 6x2 +9x 3
c. f(x) = 3
4x4 4x3 +6x2
d. f(x) = x5 +5x
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