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.
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.)
[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.
\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 $0 < c < 6$.
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.
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