For more information, please see full course syllabus of Calculus AB

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For more information, please see full course syllabus of Calculus AB

For more information, please see full course syllabus of Calculus AB

### Mean Value Theorem

- There exists a point
*c*, between*a*and*b*where the slope is defined by change in*f(x)*divided by change in*x*. - Applying Mean Value Theorem
- Check conditions
- Take derivative
- Set derivative function =
- Replace
*x*with*c* - Solve for
*c*value

### Mean Value Theorem

Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = e

^{x}on the interval [0, ln2]- f′(x) = e
^{x} - Continuous and differentiable everywhere
- f′(c) = [(f(ln2) − f(0))/(ln2 − 0)]
- f′(c) = [(e
^{ln2}− e^{0})/ln2] - f′(c) = [(2 − 1)/ln2] = [1/ln2]
- Rolle's Theorem conditions are not met (f(a) ≠ f(b)), but the Mean Value conditions are met.
- f′(c) = e
^{c}= [1/ln2] - lne
^{c}= ln[1/ln2]

c = ln[1/ln2]

Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = sinx on the interval [0, 2π]

- f′(x) = cosx
- Continuous and differentiable everywhere
- f′(c) = [(cos2π− cos0)/(2π− 0)]
- f′(c) = [(1 − 1)/(2π)]
- f′(c) = 0
- cosc = 0
- There are many solutions, but only two in the interval

c = [(π)/2], [(3π)/2]

Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = x

^{3}− x^{2}+ 1 on the interval [0, 1]- f′(x) = 3x
^{2}− 2x - f′(c) = [(1
^{3}− 1^{2}+ 1 − 0)/(1 − 0)] - f′(c) = 1
- 3c
^{2}− 2c = 1 - 3c
^{2}− 2c − 1 = 0 - c = [(2 ±√{4 − 4(3)(−1)})/(−6)]
- c = [(2 ±4)/(−6)]
- c = 1, [(−1)/3]
- Only one of these values is in the interval

c = 1

Find the mean value of f′(x) using the Mean Value Theorem for f(x) = 3x

^{2}on the interval [1, 2]- Polynomial. Continuous and differentiable.
- f′(x) = 6x
- f′(c) = [(3(2
^{2}) − 3(1^{2}))/(2 − 1)]

f′(c) = 9

Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = [1/x] on the interval [−1,1]

- f(x) is undefined at x = 0 and that is within the interval
- f(x) is not continuous on the interval

Mean Value Theorem cannot be applied

Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = [1/x] on the interval [1, 2]

- f′(x) = −[1/(x
^{2})] - f′(c) = [([1/2] − [1/1])/(2 − 1)]
- f′(c) = −[1/2]
- −[1/2] = −[1/(c
^{2})] - [1/2] = [1/(c
^{2})] - c
^{2}= 2 - c = ±√2
- Only one solution in the interval

c = √2

f(x) is continuous and differentiable on [a,b] and f(a) = f(b) and a ≠ b. Show that f′(x) has a root in the interval

- In order for some c to be a root of f′(x), f′(c) = 0
- Continuous and differentiable, so Mean Value Theorem can be used
- f′(c) = [(f(b) − f(a))/(b − a)]
- f′(c) = [(f(b) − f(b))/(b − a)]
- f′(c) = [0/(b − a)]
- f′(c) = 0

f′(c) = 0, so there is a root at x = c

Find at least one root of f′(x) on the interval [−1, 1] given f(x) = x

^{2}- Polynomial. Continuous and differentiable everywhere
- f′(x) = 2x
- f′(c) = [(1
^{2}− (−1^{2}))/(1 − (−1))] - f′(c) = 0
- 2c = 0
- c = 0

There's a root of f′(0)

f(x) is continuous and differentiable on [0,5]. Also, f(0) = 1 and f′(x) ≤ 3 in the interval. Find the largest possible value of f(5).

- f′(c) = [(f(5) − f(0))/(5 − 0)]
- f′(c) = [(f(5) − 1)/5]
- 5 f′(c) = f(5) − 1
- f(5) = 5 f′(c) + 1
- Plug in the largest value
- f(5) = 5(3) + 1

The largest possible value of f(5) is 16

Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = √x on the interval [1, 4]

- f(x) is continuous so long as x ≥ 0
- f′(x) = [1/(2 √x)]
- Differentiable so long as x > 0
- Continuous and differentiable in the interval
- f′(c) = [(√4 − √1)/(4 − 1)]
- f′(c) = [1/3]
- [1/(2 √c)] = [1/3]
- 2√c = 3
- √c = [3/2]

c = [9/4]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Mean Value Theorem

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Mean Value Theorem 0:06
- Rolle's Theorem
- Mean Value Theorem Conditions
- Mean Value Theorem Definition
- Example 1 0:56
- Example 2 2:44
- Example 3 5:28
- Example 4 7:15

0 answers

Post by Angela Zhang on December 7, 2014

John, you are very cute! I like you <3

0 answers

Post by Yu Han on July 24, 2014

For example 1, how can you tell if it met these two conditions?

0 answers

Post by Justin Powell on February 10, 2014

Just for clarity:

The Mean Value Theorem says that if a function f(x) is continuous

and differentiable between two intervals x=a and x=b,

then solving the function for these two values will give

the coordinates {a,f(a)} and {b,f(b)}.

Now if you draw a line between these two points, the slope will be:

(f(b)-f(a))/(b-a) which is the rise over the run.

The Mean Value Theoem states that at some point, {c,f(c)},

on the graph of the function between the intervals x=a an x=b,

the derivative or slope must be equal to (f(b)-f(a))/(b-a) at least once.

So the derivative of f(x) at x=c is equal to:

f'(c)=(f(b)-f(a))/(b-a)

In the last example:

the slope between the coordinates {-1,-1} and {5,125} equals

126/6 = 21=f'(c)=3c^2 c=sqrt(7).

At f(sqrt7) the slope of the curve of the graph is

parallel to the line between {-1,-1} and {5,125}.

also for the MVT, f'(c)=0 is only true when f(a) and f(b)

are equal (Rolles theorem).

Sorry for overclarifying, but the video left out some

key points.

1 answer

Last reply by: Andrew Mu

Mon Jan 6, 2014 8:30 AM

Post by Willie Wang on January 13, 2013

AT Example 4, why c cannot be negative sqr(7)ï¼Ÿ

1 answer

Last reply by: Johnny Zamora

Fri Jan 10, 2014 2:58 AM

Post by James Xie on December 17, 2012

What exactly does c stand for? Is it the Instant Rate Of Change (IROC?

0 answers

Post by Ryan Menezes on November 18, 2012

That's exactly what i want to know......

1 answer

Last reply by: Siyun Liu

Sat Jan 24, 2015 6:57 PM

Post by Steve Denton on October 18, 2012

At 5:00, what happened to the 1/2 in front of (c-1)^-1/2?

rewritten then 1/ (2 sq rt (c-1))?????