Mean Value Theorem & Rolle's Theorem - Matheno.com

Mean Value Theorem & Rolle’s Theorem

Mean Value Theorem & Rolle’s Theorem: Problems and Solutions

Are you trying to use the Mean Value Theorem or Rolle’s Theorem in Calculus? Let’s introduce the key ideas and then examine some typical problems step-by-step so you can learn to solve them routinely for yourself.

CALCULUS SUMMARY: Mean Value Theorem & Rolle's Theorem

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Mean Value Theorem (MVT):

If $f(x)$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there is a number $c$ in $(a,b)$ such that
$$\bbox[yellow,5px]{f'(c) = \frac{f(b) – f(a)}{b-a}}$$
or, equivalently,
$$\bbox[yellow,5px]{f(b) – f(a) = f'(c)(b-a)}$$

In words: there is at least one value $c$ between $a$ and $b$ where the tangent line is parallel to the secant line that connects the interval’s endpoints. (See the figures.)

When the Mean Value Theorem applies, the slope of the tangent line at x=c is the same as the slope of the secant line connecting the endpoints of the interval.

[Click on a figure to see a larger version.]

Rolle’s Theorem:
In Calculus texts and lecture, Rolle’s theorem is given first since it’s used as part of the proof for the Mean Value Theorem (MVT). You can easily remember it, though, as just a special case of the MVT: it has the same requirements about continuity on $[a,b]$ and differentiability on $(a,b)$, and the additional requirement that $f(a) = f(b)$. In that case, the MVT says that
\begin{align*}
f(b) – f(a) &= f'(c)(b-a) \\
0 &= f'(c)(b-a) \\
\end{align*}
Since $b \ne a$ (or there’s no interval), we know $b-a \ne 0.$

Hence when $f(a) = f(b),$ we must have
\[\bbox[yellow,5px]{f'(c) = 0} \] \[\text{for some number $c$ in the open interval $(a,b)$} \] In words: when $f(a) = f(b),$ the slope of the secant line connecting the endpoints is zero, and hence there is at least one value $c$ between $a$ and $b$ where the tangent line has zero slope. (See the figures.)

The problems below illustrate some typical uses of the Mean Value Theorem and Rolle’s Theorem.

Problem #1: Straightforward Application of Rolle's Theorem

Consider the function $f(x) = 9 – (x-3)^2$ on the interval $[0, 6]$. (Note that $f(0) = f(6) = 0$.) Find the value(s) of $c$ that satisfy Rolle’s Theorem.

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Since the function $f(x) = 9- (x-3)^2$ is a polynomial, it is continuous on the interval $[0, 6]$ and differentiable on the interval $(0,6)$. Furthermore, as the question states, $f(0) = f(6)$, and so Rolle’s Theorem applies. We’re looking for a value of $c$ such that $f'(c) = 0$:
\begin{align*}
f'(x) &= \frac{d}{dx}\left[9- (x-3)^2 \right] \\[8px] &= -2(x-3)
\end{align*}
We want to find $c$ such that $f'(c) = 0$:
\begin{align*}
f'(c) &= -2(c-3) = 0 \\[8px] c &= 3 \quad \cmark
\end{align*}
The point $c = 3$ has a horizontal tangent line, satisfying Rolle’s Theorem since $0 < c < 6$.

Problem #2: Straightforward Application of the Mean Value Theorem

Consider the function $f(x) = x^2$ on the interval $[1, 4]$. Find the value(s) of $c$ that satisfy the Mean-Value Theorem.

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Since the function $f(x) = x^2$ is a polynomial, it is continuous on the interval $[1, 4]$ and differentiable on the interval $(1,4)$, and so the Mean-Value Theorem applies.
We’re looking for a value of $c$ such that
\begin{align*}
f'(c) &= \frac{f(4) – f(1)}{4-1} \\[8px] &= \frac{16 – 1}{3} \\[8px] &= \frac{15}{3} = 5
\end{align*}
Now we can compute $f'(x)$ starting with $f(x) = x^2$:
\begin{align*}
f'(x) &= \frac{d}{dx}x^2 \\[8px] &= 2x
\end{align*}
We want to find $c$ such that $f'(c) = 5$:
\begin{align*}
f'(c) &= 2c = 5 \\[8px] c &= 2.5 \quad \cmark
\end{align*}
The tangent line at the point $c = 2.5$ is parallel to the secant line that connects the endpoints of the interval.

Problem #3: Prove if f '(x) > 0, then f(x) is an increasing function

Use the Mean Value Theorem to prove that if $f(x)$ is differentiable and $f'(x) > 0$ for all $x$, then $f(x)$ is an increasing function.

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This question might seem silly: “Prove that if a function has positive slope, then it is increasing.” But mathematics is partly about using theorems you’ve already proven to prove other things, no matter how obvious they may seem, so let’s just do as the question asks.

First, note that since f(x) is differentiable for all x, it must be continuous for all x, and so the Mean Value Theorem (MVT) applies. The problems says to use the MVT, so let’s start there, and consider an interval $(a,b)$. The MVT tells us that there exists a c, $a < c < b$, such that $$f(b) - f(a) = f'(c) (b-a)$$ Since $f'(x) > 0$ for all $x$, we have $f'(c) > 0$.

Furthermore, since $b > a$, $b-a > 0$.

Hence
$$f(b) – f(a) > 0$$
Hence
$$f(b) > f(a)$$
That is, $f(x)$ is increasing. $\quad \cmark$

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