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The Mean Value Theorem

Explain the meaning of Rolle’s theorem.

Rolle’s Theorem

Informally, Rolle’s theorem states that if the outputs of a differentiable function \(f\) are equal at the endpoints of an interval, then there must be an interior point \(c\) where \(f'(c)=0.\) illustrates this theorem.

Condensed — the full section is in OpenStax Calculus Volume 1.

The Mean Value Theorem and Its Meaning

Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions \(f\) defined on a closed interval \([a,b]\) with \(f(a)=f(b)\). The Mean Value Theorem generalizes Rolle’s theorem by considering functions that do not necessarily have equal value at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s theorem (). The Mean Value Theorem states that if \(f\) is continuous over the closed interval \([a,b]\) and differentiable over the open interval \((a,b),\) then there exists a point \(c\in (a,b)\) such that the tangent line to the graph of \(f\) at \(c\) is parallel to the secant line connecting \((a,f(a))\) and \((b,f(b)).\)

Condensed — the full section is in OpenStax Calculus Volume 1.

Corollaries of the Mean Value Theorem

Let’s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use in upcoming sections.

At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to conclude that the converse is also true. In particular, if \({f}^{'}(x)=0\) for all \(x\) in some interval \(I,\) then \(f(x)\) is constant over that interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss them shortly.

Since \(f\) is differentiable over \(I,\) \(f\) must be continuous over \(I.\) Suppose \(f(x)\) is not constant for all \(x\) in \(I.\) Then there exist \(a,b\in I,\) where \(a\ne b\) and \(f(a)\ne f(b).\) Choose the notation so that \(a\[\frac{f(b)-f(a)}{b-a}\ne 0.\]

Since \(f\) is a differentiable function, by the Mean Value Theorem, there exists \(c\in (a,b)\) such that

\[{f}^{'}(c)=\frac{f(b)-f(a)}{b-a}.\]

Therefore, there exists \(c\in I\) such that \({f}^{'}(c)\ne 0,\) which contradicts the assumption that \({f}^{'}(x)=0\) for all \(x\in I.\)

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From , it follows that if two functions have the same derivative, they differ by, at most, a constant.

Let \(h(x)=f(x)-g(x).\) Then, \({h}^{'}(x)={f}^{'}(x)-{g}^{'}(x)=0\) for all \(x\in I.\) By Corollary 1, there is a constant \(C\) such that \(h(x)=C\) for all \(x\in I.\) Therefore, \(f(x)=g(x)+C\) for all \(x\in I.\)

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This fact is important because it means that for a given function \(f,\) if there exists a function \(F\) such that \({F}^{'}(x)=f(x);\) then, the only other functions that have a derivative equal to \(f\) are \(F(x)+C\) for some constant \(C.\) We discuss this result in more detail later in the chapter.

The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall that a function \(f\) is increasing over \(I\) if \(f({x}_{1})f({x}_{2})\) whenever \({x}_{1}<{x}_{2}.\) Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing (). We make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum and minimum values of the function, and how to determine the shape of the graph.

Condensed — the full section is in OpenStax Calculus Volume 1.

Key Concepts

If \(f\) is continuous over \([a,b]\) and differentiable over \((a,b)\) and \(f(a)=f(b),\) then there exists a point \(c\in (a,b)\) such that \({f}^{'}(c)=0.\) This is Rolle’s theorem.
If \(f\) is continuous over \([a,b]\) and differentiable over \((a,b),\) then there exists a point \(c\in (a,b)\) such that
\[f'(c)=\frac{f(b)-f(a)}{b-a}.\]
This is the Mean Value Theorem.
If \(f'(x)=0\) over an interval \(I,\) then \(f\) is constant over \(I.\)
If two differentiable functions \(f\) and \(g\) satisfy \({f}^{'}(x)={g}^{'}(x)\) over \(I,\) then \(f(x)=g(x)+C\) for some constant \(C.\)
If \({f}^{'}(x)>0\) over an interval \(I,\) then \(f\) is increasing over \(I.\) If \({f}^{'}(x)<0\) over \(I,\) then \(f\) is decreasing over \(I.\)

The Mean Value Theorem

For the following exercises, determine over what intervals (if any) the Mean Value Theorem applies. Justify your answer.

For the following exercises, graph the functions on a calculator and draw the secant line that connects the endpoints. Estimate the number of points \(c\) such that \({f}^{'}(c)(b-a)=f(b)-f(a).\)

For the following exercises, find all points \(0

For the following exercises, show there is no \(c\) such that \(f(1)-f(-1)={f}^{'}(c)(2).\) Explain why the Mean Value Theorem does not apply over the interval \([-1,1].\)

For the following exercises, determine whether the Mean Value Theorem applies for the functions over the given interval \([a,b].\) Justify your answer.

For the following exercises, consider the roots of the equation.

For the following exercises, use a calculator to graph the function over the interval \([a,b]\) and graph the secant line from \(a\) to \(b.\) Use the calculator to estimate all values of \(c\) as guaranteed by the Mean Value Theorem. Then, find the exact value of \(c,\) if possible, or write the final equation and use a calculator to estimate to four digits.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

For each of the following functions, verify that the function satisfies the criteria stated in Rolle’s theorem and find all values \(c\) in the given interval where \(f'(c)=0.\)
1. \(f(x)={x}^{2}+2x\) over \([-2,0]\)
2. \(f(x)={x}^{3}-4x\) over \([-2,2]\)
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1. Since \(f\) is a polynomial, it is continuous and differentiable everywhere. In addition, \(f(-2)=0=f(0).\) Therefore, \(f\) satisfies the criteria of Rolle’s theorem. We conclude that there exists at least one value \(c\in (-2,0)\) such that \(f'(c)=0.\) Since \(f'(x)=2x+2=2(x+1),\) we see that \(f'(c)=2(c+1)=0\) implies \(c=-1\) as shown in the following graph.
2. As in part a. \(f\) is a polynomial and therefore is continuous and differentiable everywhere. Also, \(f(-2)=0=f(2).\) That said, \(f\) satisfies the criteria of Rolle’s theorem. Differentiating, we find that \(f'(x)=3{x}^{2}-4.\) Therefore, \(f'(c)=0\) when \(x=\text{\pm }\frac{2}{\sqrt{3}}.\) Both points are in the interval \([-2,2],\) and, therefore, both points satisfy the conclusion of Rolle’s theorem as shown in the following graph.
Verify that the function \(f(x)=2{x}^{2}-8x+6\) defined over the interval \([1,3]\) satisfies the conditions of Rolle’s theorem. Find all points \(c\) guaranteed by Rolle’s theorem.

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\(c=2\)
For \(f(x)=\sqrt{x}\) over the interval \([0,9],\) show that \(f\) satisfies the hypothesis of the Mean Value Theorem, and therefore there exists at least one value \(c\in (0,9)\) such that \({f}^{'}(c)\) is equal to the slope of the line connecting \((0,f(0))\) and \((9,f(9)).\) Find these values \(c\) guaranteed by the Mean Value Theorem.

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We know that \(f(x)=\sqrt{x}\) is continuous over \([0,9]\) and differentiable over \((0,9).\) Therefore, \(f\) satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value \(c\in (0,9)\) such that \({f}^{'}(c)\) is equal to the slope of the line connecting \((0,f(0))\) and \((9,f(9))\) (). To determine which value(s) of \(c\) are guaranteed, first calculate the derivative of \(f.\) The derivative \({f}^{'}(x)=\frac{1}{(2\sqrt{x})}.\) The slope of the line connecting \((0,f(0))\) and \((9,f(9))\) is given by
\[\frac{f(9)-f(0)}{9-0}=\frac{\sqrt{9}-\sqrt{0}}{9-0}=\frac{3}{9}=\frac{1}{3}.\]
We want to find \(c\) such that \({f}^{'}(c)=\frac{1}{3}.\) That is, we want to find \(c\) such that
\[\frac{1}{2\sqrt{c}}=\frac{1}{3}.\]
Solving this equation for \(c,\) we obtain \(c=\frac{9}{4}.\) At this point, the slope of the tangent line equals the slope of the line joining the endpoints.
If a rock is dropped from a height of 100 ft, its position \(t\) seconds after it is dropped until it hits the ground is given by the function \(s(t)=-16{t}^{2}+100.\)
1. Determine how long it takes before the rock hits the ground.
2. Find the average velocity \({v}_{\text{avg}}\) of the rock for when the rock is released and the rock hits the ground.
3. Find the time \(t\) guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is \({v}_{\text{avg}}.\)
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1. When the rock hits the ground, its position is \(s(t)=0.\) Solving the equation \(-16{t}^{2}+100=0\) for \(t,\) we find that \(t=\text{\pm }\frac{5}{2}\ \text{sec}.\) Since we are only considering \(t\ge 0,\) the ball will hit the ground \(\frac{5}{2}\) sec after it is dropped.
2. The average velocity is given by
  \[{v}_{\text{avg}}=\frac{s(5\text{/}2)-s(0)}{5\text{/}2-0}=\frac{0-100}{5\text{/}2}=-40\ \text{ft/sec}.\]
3. The instantaneous velocity is given by the derivative of the position function. Therefore, we need to find a time \(t\) such that \(v(t)={s}^{'}(t)={v}_{\text{avg}}=-40\ \text{ft/sec}.\) Since \(s(t)\) is continuous over the interval \([0,5\text{/}2]\) and differentiable over the interval \((0,5\text{/}2),\) by the Mean Value Theorem, there is guaranteed to be a point \(c\in (0,5\text{/}2)\) such that
  \[{s}^{'}(c)=\frac{s(5\text{/}2)-s(0)}{5\text{/}2-0}=-40.\]
  Taking the derivative of the position function \(s(t),\) we find that \({s}^{'}(t)=-32t.\) Therefore, the equation reduces to \({s}^{'}(c)=-32c=-40.\) Solving this equation for \(c,\) we have \(c=\frac{5}{4}.\) Therefore, \(\frac{5}{4}\) sec after the rock is dropped, the instantaneous velocity equals the average velocity of the rock during its free fall: \(-40\) ft/sec.
Suppose a ball is dropped from a height of 200 ft. Its position at time \(t\) is \(s(t)=-16{t}^{2}+200.\) Find the time \(t\) when the instantaneous velocity of the ball equals its average velocity.

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\(\frac{5}{2\sqrt{2}}\) sec
Why do you need continuity to apply the Mean Value Theorem? Construct a counterexample.
Why do you need differentiability to apply the Mean Value Theorem? Find a counterexample.

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One example is \(f(x)=|x|+3,\ -2\le x\le 2\)
When are Rolle’s theorem and the Mean Value Theorem equivalent?
If you have a function with a discontinuity, is it still possible to have \({f}^{'}(c)(b-a)=f(b)-f(a)?\) Draw such an example or prove why not.

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Yes, but the Mean Value Theorem still does not apply
\(y=\text{sin}(\pi x)\)
\(y=\frac{1}{{x}^{3}}\)

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Any closed interval in \((\text{-}\infty ,0)\text{or}(0,\infty )\)
\(y=\sqrt{4-{x}^{2}}\)
\(y=\sqrt{{x}^{2}-4}\)

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Any closed interval in \((\text{-}\infty ,-2)\text{or}(2,\infty )\)
\(y=\text{ln}(3x-5)\)
[T] \(y=3{x}^{3}+2x+1\) over \([-1,1]\)

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2 points
[T] \(y=\text{tan}(\frac{\pi }{4}x)\) over \([-\frac{3}{2},\frac{3}{2}]\)
[T] \(y={x}^{2}\text{cos}(\pi x)\) over \([-2,2]\)

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5 points
[T] \(y={x}^{6}-\frac{3}{4}{x}^{5}-\frac{9}{8}{x}^{4}+\frac{15}{16}{x}^{3}+\frac{3}{32}{x}^{2}+\frac{3}{16}x+\frac{1}{32}\) over \([-1,1]\)
\(f(x)={x}^{3}\)

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\(c=\frac{2\sqrt{3}}{3}\)
\(f(x)=\text{sin}(\pi x)\)
\(f(x)=\text{cos}(2\pi x)\)

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\(c=\frac{1}{2},1,\frac{3}{2}\)
\(f(x)=1+x+{x}^{2}\)
\(f(x)={(x-1)}^{10}\)

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\(c=1\)
\(f(x)={(x-1)}^{9}\)
\(f(x)=|x-\frac{1}{2}|\)

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Not differentiable
\(f(x)=\frac{1}{{x}^{2}}\)
\(f(x)=\sqrt{|x|}\)

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Not differentiable
\(f(x)=⌊x⌋\) (Hint: This is called the floor function and it is defined so that \(f(x)\) is the largest integer less than or equal to \(x.)\)
\(y={e}^{x}\) over \([0,1]\)

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Yes
\(y=\text{ln}(2x+3)\) over \([-\frac{3}{2},0]\)
\(f(x)=\text{tan}(2\pi x)\) over \([0,2]\)

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The Mean Value Theorem does not apply since the function is discontinuous at \(x=\frac{1}{4},\frac{3}{4},\frac{5}{4},\frac{7}{4}.\)
\(y=\sqrt{9-{x}^{2}}\) over \([-3,3]\)
\(y=\frac{1}{|x+1|}\) over \([0,3]\)

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Yes
\(y={x}^{3}+2x+1\) over \([0,6]\)
\(y=\frac{{x}^{2}+3x+2}{x}\) over \([-1,1]\)

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The Mean Value Theorem does not apply; discontinuous at \(x=0.\)
\(y=\frac{x}{\text{sin}(\pi x)+1}\) over \([0,1]\)
\(y=\text{ln}(x+1)\) over \([0,e-1]\)

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Yes
\(y=x\ \text{sin}(\pi x)\) over \([0,2]\)
\(y=5+|x|\) over \([-1,1]\)

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The Mean Value Theorem does not apply; not differentiable at \(x=0.\)
Show that the equation \(y={x}^{3}+4x+16\) has exactly one real root. What is it?

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

\neq

not equal
The two sides are different.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: The Mean Value Theorem

Explain the meaning of Rolle’s theorem.
Describe the significance of the Mean Value Theorem.
State three important consequences of the Mean Value Theorem.
There exists
There exists
Since
As in part a.
Determine how long it takes before the rock hits the ground.

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

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Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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