# Mean Value Theorem

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The Mean Value Theorem – HMC Calculus Tutorial

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.

#### 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 $$f'(c) = \frac{f(b)-f(a)}{b-a}.$$