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For more information, please see full course syllabus of College Calculus: Level I

For more information, please see full course syllabus of College Calculus: Level I

### Mean Value Theorem and Rolle's Theorem

- You should memorize the Mean Value Theorem and Rolle’s Theorem – including the continuity and differentiability hypotheses.
- If you are using one of these theorems, do check that the continuity and differentiability hypotheses are satisfied.
- If a problem asks you to verify a conclusion of one of these theorems, use algebra to do so.
- Note that Rolle’s Theorem is simply a special case of the Mean Value Theorem.

### Mean Value Theorem and Rolle's Theorem

Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [0, 4] with f(x) = 4x

^{2}− 16x + 4- Polynomial, so it's continuous
- f′(x) = 8x − 16 defined for all x
- 4x
^{2}− 16x + 4 = 0 - x = [(16 ±√{(−16)
^{2}− 4(4)(4)})/2(4)] - x = [(16 ±√{256 − 64})/8]
- x = 2 ±[(√{192})/8]
- x = 2 ±[(√{64 * 3})/8]
- x = 2 ±√3 Both solutions are within the interval

Yes. There exists some c in the interval [0,4] such that f′(c) = 0.

Find the value of x such that f′(x) = 0 for f(x) = 4x

^{2}− 16x + 4- f′(x) = 8x − 16
- 8x − 16 = 0
- 8x = 16
- x = [16/8]

f′(2) = 0

Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [−1, 1] with f(x) = |x| − 1

- We can rewrite this as f(x) = √{x
^{2}} − 1 - The function is continuous on the interval
- f′(x) = [1/2] [1/(√{x
^{2}})] [d/dx] x^{2} - f′(x) = [x/(√{x
^{2}})] Removable discontinuity at x = 0 - lim
_{x → 0+}f′(x) ≠ lim_{x → 0−}f′(x)

The function is not differentiable at x = 0, which is in the interval. Rolle's Theorem cannot be applied.

Confirm the Rolle's Theorem conditions on the interval [−1, 1] with f(x) = x − 1

- Polynomial, so it's continuous
- f′(x) = 1
- Differentiable on that interval
- Let's check the endpoints
- f(−1) = −2
- f(1) = 0
- The endpoints don't match. They don't necessarily have to equal zero, but there must be two equal values of y within that interval. Rolle's Theorem is only a special case of the Mean Value Theorem, which is covered in the next lesson

The conditions for Rolle's Theorem are not met.

Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [0, π] with f(x) = sinx

- f(x) = sinx
- sinx is defined for all x. Continuous
- f′(x) = cosx
- cosx is defined for all x and has no edges or kinks in its graph. Differentiable.
- sinx = 0
- x = 0, π

Yes. There exists some c in the interval [0, π] such that f′(c) = 0.

Find the value of x in the interval [0, π] such that f′(x) = 0 for f(x) = sinx

- f′(x) = cosx
- f′([(π)/2]) = 0

f′([(π)/2]) = 0

Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [0, 4] with f(x) = x

^{3}− 12x- Polynomial function, so it's continuous everywhere
- f′(x) = 3x
^{2}− 12 continuous everywhere - x
^{3}− 12x = 0 - x(x
^{2}− 12) = 0 - x(x + √{12})(x − √{12}) = 0
- Three solutions, x = 0, −√{12}, √{12}, and two of these are within the interval

Yes. There exists some c in the interval [0, 4] such that f′(c) = 0.

Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [−2, 2] with f(x) = e

^{x}- e
^{x}is a continuous function so it's continuous function. In fact, it can be expressed as a polynomial function - f′(x) = e
^{x}defined everywhere. In fact, it can be expressed as a polynomial function - e
^{x}≠ 0 - Let's test the end-points
- f(−2) = e
^{−2} - f(2) = e
^{2} - f(2) ≠ f(−2)

Rolle's Theorem conditions not met.

Find x such that f′(x) = 0 where f(x) = x

^{2}- f′(x) = 2x
- f′(0) = 0

When x = 0, f′(x) = 0. Rolle's Theorem can be used to prove that a solution in an interval exists, but it doesn't necessarily prove there is no solution. In truth, the same

Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [0, 1] with f(x) = x

^{4}− x^{2}- Polynomial function, so it's continuous
- f′(x) = 4x
^{3}− 2x is defined everywhere - x
^{4}− x^{2}= 0 - x
^{2}(x^{2}− 1) = 0 - x = −1, 0, 1

Yes. There exists some c in the interval [0, 1] such that f′(c) = 0.

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]

Find the average value, f(c), of of the function f(x) = sinx on the interval [0, π] using the Mean Value Theorem for Integrals

- f(c) = [1/(π− 0)] ∫
_{0}^{π}sinx dx - f(c) = [1/(π)] (−cosx) |
_{0}^{π} - f(c) = [1/(π)] (− cosπ+ cos0)
- f(c) = [1/(π)] (1 + 1)

f(c) = [2/(π)]

Find the average value, f(c) of of the function f(x) = sinx on the interval [0, 2π] using the Mean Value Theorem for Integrals

- f(c) = [1/(2π− 0)] ∫
_{0}^{2π}sinx dx - f(c) = [1/(2π)] (−cosx) |
_{0}^{2π} - f(c) = [1/(2π)] (− cos2π+ cos0)
- f(c) = [1/(2π)] (−1 + 1)

f(c) = 0

Find the average value,f(c), of of the function f(x) = x

^{2}+ 7 on the interval [0, 3] using the Mean Value Theorem for Integrals- f(c) = [1/(3 − 0)] ∫
_{0}^{3}x^{2}+ 7 dx - f(c) = [1/3] ([(x
^{3})/3] + 7x) |_{0}^{3} - f(c) = [1/3] ([(3
^{3})/3] + 21 − (0 + 0)) - f(c) = [1/3] 30

f(c) = 10

Find the average value,f(c), of of the function f(x) = x

^{3}+ 2x + 1 on the interval [0, 4] using the Mean Value Theorem for Integrals- f(c) = [1/(4 − 0)] ∫
_{0}^{4}x^{3}+ 2x + 1 dx - f(c) = [1/4] ([(x
^{4})/4] + x^{2}+ x) |_{0}^{4} - f(c) = [1/4] ([(4
^{4})/4] + 4^{2}+ 4 − 0 − 0 − 0) - f(c) = 4
^{2}+ 4 + 1

f(c) = 21

Find the average value,f(c), of of the function f(x) = [1/x] on the interval [−1, 1] using the Mean Value Theorem for Integrals

- f(x) has a discontinuity at x = 0 and that is in the interval

Mean Value Theorem for Integrals cannot be applied

Find the average value,f(c), of of the function f(x) = [1/x] on the interval [1, e] using the Mean Value Theorem for Integrals

- f(x) is continuous for x > 0
- f(c) = [1/(e − 1)] ∫
_{1}^{e}[1/x] dx - f(c) = [1/(e − 1)] lnx |
_{1}^{e} - f(c) = [1/(e − 1)] (lne − ln1)
- f(c) = [1/(e − 1)] (1)

f(c) = [1/(e − 1)]

Find the average value,f(c), of of the function f(x) = cosx + 4 on the interval [0, 2π] using the Mean Value Theorem for Integrals

- f(c) = [1/(2π− 0)] ∫
_{0}^{2π}cosx + 4 dx - f(c) = [1/(2π)] (sinx + 4x) |
_{0}^{2π} - f(c) = [1/(2π)] (sin2π+ 8 π− sin0 − 0)
- f(c) = [1/(2π)] (8 π)

f(c) = 4

Find the average value,f(c), of of the function f(x) = [1/(√{4 − x

^{2}})] on the interval [0, 1] using the Mean Value Theorem for Integrals- f(x) is undefined for x ≤ −2 and x ≥ 2
- The function is defined on the interval [0, 1]
- f(c) = [1/(1 − 0)] ∫
_{0}^{1}[1/(√{4 − x^{2}})] dx - f(c) = ∫
_{0}^{1}[1/(√{2^{2}− x^{2}})] dx - f(c) = sin
^{−1}[x/2] |_{0}^{1} - f(c) = sin
^{−1}[1/2] − sin^{−1}0 - f(c) = [(π)/6] − 0

f(c) = 0

Find the average value,f(c), of of the function f(x) = 4x on the interval [−5, 5] using the Mean Value Theorem for Integrals

- f(c) = [1/(5 − (−5))] ∫
_{−5}^{5}4x dx - f(c) = [1/10] 2x
^{2}|_{−5}^{5} - f(c) = [1/10] (2(5)
^{2}− 2(−5)^{2}) - f(c) = [1/10] (0)

f(c) = 0

Find the average value,f(c), of of the function f(x) = tan[x/4] on the interval [0, π] using the Mean Value Theorem for Integrals

- f(x) has discontinuities whenever cos[x/4] = 0
- So near zero, f(x) is discontinuous at −2π and 2π. Our interval does not include a discontinuity
- f(c) = [1/(π− 0)] ∫
_{0}^{π}tan[x/4] dx - u = [x/4]
- du = [1/4] dx
- f(c) = [1/(π)] 4 ∫
_{0}^{π}tanu du - f(c) = [4/(π)] ∫
_{0}^{π}[sinu/cosu] du - v = cosu
- dv = −sinu du
- f(c) = [4/(π)] (−1) ∫
_{0}^{π}[1/v] dv - f(c) = − [4/(π)] lnv |
_{0}^{π} - f(c) = − [4/(π)] lncosu |
_{0}^{π} - f(c) = − [4/(π)] lncos[x/4] |
_{0}^{π} - f(c) = − [4/(π)] (lncos[(π)/4] − lncos0)
- f(c) = − [4/(π)] (ln[1/(√2)] − ln1)
- f(c) = − [4/(π)] (ln[1/(√2)] − 0)
- f(c) = − [4/(π)] (ln1 − ln√2)
- f(c) = − [4/(π)] (0 − ln2
^{[1/2]}) - f(c) = − [4/(π)] (− [1/2] ln2)

f(c) = [2 ln2/(π)]

*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 and Rolle's 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
- Theorems 0:09
- Mean Value Theorem
- Graphical Explanation
- Rolle's Theorem
- Graphical Explanation
- Lecture Example 1 3:36
- Lecture Example 2 6:33
- Lecture Example 3 9:32
- Additional Example 4
- Additional Example 5

1 answer

Last reply by: Cheng Jiang

Wed Oct 30, 2013 5:37 PM

Post by Shehan Gunasekara on May 1, 2012

Thanks!! helped alot with uni

0 answers

Post by Real Schiran on October 29, 2011

:-)

0 answers

Post by Ahmed Shiran on October 29, 2011

Cool !