Mean Value Theorem

The Mean Value Theorem – HMC Calculus Tutorial

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We begin with a common-sense geometrical fact:

somewhere between two zeros of a non-constant continuous function $f$,the function must change direction

For a differentiable function, the derivative is $0$ at the point where $f$ changes direction. Thus, we expect there to be a point $c$ where the tangent is horizontal. These ideas are precisely stated by Rolle’s Theorem:

Rolle’s Theorem

Let $f$ be differentiable on $(a,b)$ and continuous on $[a,b]$. If $f(a)=f(b)=0$, then there is at least one point $c$ in $(a,b)$ for which $f'(c)=0$.

Notice that both conditions on $f$ are necessary. Without either one, the statement is false!

For a discontinuous function, the conclusion of Rolle’s Theorem may not hold:

For a continuous, non-differentiable function, again this might not be the case:

Though the theorem seems logical, we cannot be sure that it is always true without a proof.

Note that either $f(x)$ is always $0$ on $[a,b]$ or $f$ varies on $[a,b]$.

If $f(x)$ is always $0$, then $f'(x)=0$ for all $x$ in $(a,b)$ and we are done.
If $f(x)$ varies on $(a,b)$, then there must be points where $f(x) > 0$ or points where $f(x) < 0$.
Assume first that there are points where $f(x) > 0$. By the Value Theorem $f$ has a maximum at some point $c$ in $(a,b)$. Then $f(c) > 0$, so $c$ is not an endpoint. At this maximum, $f'(c)=0$. Now assume that there are points where $f(x) < 0$. Then, again by the Value Theorem, $f$ has a minimum at some point $c$ in $(a,b)$. Again, $c$ is not an endpoint since $f(c) < 0$ while $f(a)=f(b)=0$. At this minimum, $f'(c)=0$.

This completes the proof.

The Mean Value Theorem is a generalization of Rolle’s Theorem:

We now let $f(a)$ and $f(b)$ have values other than $0$ and look at the secant line through $(a,f(a))$ and $(b,f(b))$. We expect that somewhere between $a$ and $b$ there is a point $c$ where the tangent is parallel to this secant.

In Rolle’s Theorem, the secant was horizontal so we looked for a horizontal tangent.

That is, the slopes of these two lines are equal. This is formalized in the Mean Value Theorem.

Let $f$ be differentiable on $(a,b)$ and continuous on $[a,b]$. Then there is at least one point $c$ in $(a,b)$ for which \[f'(c)=\frac{f(b)-f(a)}{b-a}.\]

Here, $f'(c)$ is the slope of the tangent at $c$, while $\displaystyle \frac{f(b)-f(a)}{b-a}$ is the slope of the secant through $a$ and $b$. Intuitively, we see that if we translate the secant line in the figure upwards, it will eventually just touch the curve at the single point $c$ and will be tangent at $c$. However, basing conclusions on a single example can be disastrous, so we need a proof.

The equation of the secant through $(a,f(a))$ and $(b,f(b))$ is \[y-f(a)=\frac{f(b)-f(a)}{b-a}(x-a)\] which we can rewrite as \[y=\frac{f(b)-f(a)}{b-a}(x-a)+f(a).\] Let \[g(x)=f(x)-\left[\frac{f(b)-f(a)}{b-a}(x-a)+f(a)\right].\] Note that $g(a)=g(b)=0$. Also, $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$ since $f$ is. So by Rolle’s Theorem there exists $c$ in $(a,b)$ such that $g'(c)=0$.

But $\displaystyle g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}$, so \[g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}=0.\] Therefore, \[f'(c)=\frac{f(b)-f(a)}{b-a}\] and the proof is complete.

Consequences of the Mean Value Theorem

The Mean Value Theorem is behind many of the important results in calculus. The following statements, in which we assume $f$ is differentiable on an open interval $I$, are consequences of the Mean Value Theorem:

$f'(x)=0$ everywhere on $I$ if and only if $f$ is constant on $I$.
If $f'(x)=g'(x)$ for all $x$ on $I$, then $f$ and $g$ differ at most by a constant on $I$.
If $f'(x)>0$ for all $x$ on $I$, then $f$ is increasing on $I$.
If $f'(x) < 0$ for all $x$ on $I$, then $f$ is decreasing on $I$.

Key Concepts

Mean Value Theorem

Let $f$ be differentiable on $(a,b)$ and continuous on $[a,b]$. Then there is at least one point $c$ in $(a,b)$ for which $$f'(c) = \frac{f(b)-f(a)}{b-a}.$$

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