# The Unapologetic Mathematician

## Consequences of the Mean Value Theorem

So now that we have the mean value theorem what can we do with it? First off, we can tell something that seems intuitively obvious. We know that a constant function has the constant zero function as its derivative. It turns out that these are the only functions with zero derivative.

To see this, let $f$ be a differentiable function on $(a,b)$ so that $f'(x)=0$ for all $x\in(a,b)$. Let $x_1$ and $x_2$ be any points between $a$ and $b$ with $x_1. Then $f$ restricts to a continuous function on the interval $\left[x_1,x_2\right]$ which is differentiable on the interior $(x_1,x_2)$. The differentiable mean value theorem then applies, and it tells us that there is some $c\in(x_1,x_2)$ with $f'(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}$. But by assumption this derivative is zero, and so $f(x_2)=f(x_1)$. Since the points were arbitrary, $f$ takes the same value at each point in $(a,b)$.

What about a function $f$ for which $f'(x)>0$ on some interval $(a,b)$? Looking at the graph it seems that the slope of all the tangent lines should be positive, and so the function should be increasing. Indeed this is the case.

Specifically we have to show that if $x_2>x_1$ for two points in $(a,b)$ then $f(x_2)>f(x_1)$. Again we look at the restriction of $f$ to a continuous function on $\left[x_1,x_2\right]$ which is differentiable on $(x_1,x_2)$. Then the mean value theorem tells us that there is some $c\in(x_1,x_2)\subseteq(a,b)$ with $f'(x)=\frac{f(x_2)-f(x_1)}{x_2-x_1}$. By assumption this quantity is positive, as is $x_2-x_1$, and so $f(x_2)>f(x_1)$. Similarly we can show that if $f'(x)<0$ on an interval $(a,b)$ then the function is decreasing there.

January 24, 2008 Posted by | Analysis, Calculus | 14 Comments