• Calculus Date: 12/10/13

Obj: SWBAT apply Rolle’s and the Mean Value Thmhttp://youtu.be/4j2ZLtGoiLE derivatives of absolute values

http://youtu.be/PBKnttVMbV4 first derivative test inc. dec.

Do Now –

HW Requests: pg 198 #11, 12, 13, 15, 17, 22, 23, 25, 29, 39, 41

SM pg 103 Read Section 5.2• pg 183 #1-21 odds• In class: page SM 82/83• HW: pg SM 84/85• Step by Step Manual pg 4, 5, 8, 11-Friday• In class: SM 102• Announcements:• No class next Wednesday• Step by Step Assignment Friday

"Do not judge me by my successes, judge me by how many times

I fell down and got back up again.“Nelson Mandela

The Mean Value Theorem

Lesson 4.2

I wonder how mean this

theorem really is?

Think About It

• Consider a trip of two hours that is 120 miles in distance … You have averaged 60 miles per hour

• What reading on your speedometer would you have expected to see at least once?

60

Rolle’s Theorem

• Given f(x) on closed interval [a, b] Differentiable on open interval (a, b)

• If f(a) = f(b) … then There exists at least one number

a < c < b such that f ’(c) = 0

f(a) = f(b)

a bc

Ex. Find the two x-intercepts of f(x) = x2 – 3x + 2and show that f’(x) = 0 at some point between the

two intercepts.

f(x) = x2 – 3x + 20 = (x – 2)(x – 1)x-int. are 1 and 2

f’(x) = 2x - 3

0 = 2x - 3

x = 3/2

Rolles Theorem is satisfied as there is a point atx = 3/2 where f’(x) = 0.

Let f(x) = x4 – 2x2 . Find all c in the interval (-2, 2)such that f’(x) = 0.

Since f(-2) and f(2) = 8, we can use Rolle’s Theorem.

f’(x) = 4x3 – 4x = 0

4x(x2 – 1) = 0

x = -1, 0, and 1

Thus, in the interval(-2, 2), the derivative

is zero at each of thesethree x-values.

8

Mean Value Theorem• We can “tilt” the picture of Rolle’s Theorem

Stipulating that f(a) ≠ f(b)

• Then there exists a c such that

a bc

( ) ( )'( )

f b f af c

b a

The Mean Value Theorem (MVT)aka the ‘crooked’ Rolle’s Theorem

If f is continuous on [a, b] and differentiable on (a, b)

There is at least one number c on (a, b) at which

ab

f(a)

f(b)

c

Conclusion:Slope of Secant Line

EqualsSlope of Tangent Line

ab

afbfcf

)()(

)('

We can “tilt” the picture of Rolle’s TheoremStipulating that f(a) ≠ f(b)

Finding c

• Given a function f(x) = 2x3 – x2 Find all points on the interval [0, 2] where

• Strategy Find slope of line from f(0) to f(2) Find f ‘(x) Set f ‘(x) equal to slope … solve for x

( ) ( )'( )

f b f af c

b a

Given f(x) = 5 – 4/x, find all c in the interval (1,4)such that the slope of the secant line = the slope of

the tangent line.

?

14

)1()4()('

ff

cf 114

14

?)(' xf2

4

x

14

2

x24 x2x

But in the interval of (1,4),only 2 works, so c = 2.

Find the value(s) of c that satisfy the Mean Value Theorem for

1f x x on 4, 4

x

1 17 1 17f 4 4 f 4 4

4 4 4 4

17 17f b f a 174 4

b a 4 4 16

2

17 1f ' c 1

16 x

2 f b f aIf f x x 2x 1, a 0, b 1, and f ' c , find c.

b a

f(0) = -1 f(1) = 2

f b f a 2 13

b a 1 0

f ' x 2x 2

3 2x 2 1

x2

f(3) = 39 f(-2) = 64

f b f a 64 395

b a 2 3

For how many value(s) of c is f ‘ (c ) = -5?

If , how many numbers on [-2, 3] satisfythe conclusion of the Mean Value Theorem.

2 2f x x 12 x 4

A. 0 B. 1 C. 2 D. 3 E. 4

CALCULATOR REQUIRED

X X X

Find the value(s) of c that satisfy the Mean Value Theorem for

1f x x on 4, 4

x

Note: The Mean Value Theorem requires the function to be continuous on [-4, 4] and differentiable on (-4, 4). Therefore, sincef(x) is discontinuous at x = 0 which is on [-4, 4], there may be no

value of c which satisfies the Mean Value Theorem

Since has no real solution, there is no value of c on

[-4, 4] which satisfies the Mean Value Theorem

2

1 1

16 x

Mean Value Theorem

• Applied to a cubic equation

Note Geogebera Example

Note Geogebera Example

Find the value(s) of c which satisfy Rolle’s Theorem for on the interval [0, 1]. 4f x x x

Verify…..f(0) = 0 – 0 = 0 f(1) = 1 – 1 = 0

3f ' x 4x 1 30 4x 1

31

c4

which is on [0, 1]

Given the graph of f(x) below, use the graph of f to estimate thenumbers on [0, 3.5] which satisfy the conclusion of the Mean Value

Theorem.

2Determine whether f x x 2x 2 satisfies the hypothesis of

the Mean Value Theorem on -2, 2 . If it does, find all numbers

f b f ac in (a, b) such that f ' c

b a

f(x) is continuous and differentiable on [-2, 2]

f 2 f 2 6 2

2 2 42

f ' x 2x 2

2x 2 2 c 0

On the interval [-2, 2], c = 0 satisfies the conclusion of MVT

2x 1Determine whether f x satisfies the hypothesis of

x 2the Mean Value Theorem on -2, 1 . If it does, find all numbers

f b f ac in (a, b) such that f ' c

b a

f(x) is continuous and differentiable on [-2, 1]

30

f 1 f 2 41 2 3

1

4

2

2

2x x 2 1 x 1

xf ' x

2

2

2

x 4x 1 1

x 4x 4 4

2 24x 16x 4 x 4x 4 23x 12x 0

3x x 4 0 On the interval [-2, 1], c = 0 satisfies the conclusion of MVT

2x 1Determine whether f x satisfies the hypothesis of

x 2the Mean Value Theorem on 0, 4 . If it does, find all numbers

f b f ac in (a, b) such that f ' c

b a

Since f(x) is discontinuous at x = 2, which is part of the interval[0, 4], the Mean Value Theorem does not apply

3Determine whether f x x 3x 1 satisfies the hypothesis of

the Mean Value Theorem on -1, 2 . If it does, find all numbers

f b f ac in (a, b) such that f ' c

b a

f(x) is continuous and differentiable on [-1, 2]

f0

2 f 1 3 3

2 1 3

23x 3f x' 23x 3 0

c = 1 satisfies the conclusion of MVT

3 x 1 x 1 0

Modeling Problem• Two police cars are located at fixed points 6

miles apart on a long straight road. The speed limit is 55 mph A car passes the first point at 53 mph Five minutes later he passes the second at 48

mph Yuk! Yuk! I think he was

speeding, EnosWe need to

prove it, Rosco