# Verify Rolle's theorem.

The given function is $f(x) = |9-x^2|$ on $[-3,3]$; also find point where the derivative is zero.

I got the value of $c$ as $\pm3$, but according to the theorem the value should be in between $-3$ and $3$, i.e., excluding the $-3$ and $3$.

The function $|9-x^2|$ is not differentiable on $-3$ and $3$, so the derivatives cannot be zero at that point, as they do not exist at all. You must have made a mistake somewhere.

In your problem particularly, as $f(x)=9-x^2$ we have $f'(x)=-2x$, as $9-x^2 \ge 0$ on $[-3,3]$, the point where the derivative is $0$ is $0$, and only $0$.

Also, note that Rolle's Theorem says that the point where the derivatives are zero are all in $[a,b]$ where $f(a)=f(b)$. It merely says that there exists such a point.

For example, take $f(x)=0$ on $[1,3]$. There exists a value where $f'(x)=0$ on $(1,3)$, but $f'(1)=f'(3)=0$ as well. So values where $f'(x)=0$ need not always exist in the interval $(a,b)$.

• I probably differentiated it wrong. I took the given function as y and then squared both sides this removing the modulus sign. Then, after differentiation I substituted the value of y. Mar 6, 2017 at 6:57
• @shoaibashraf Yes, you probably made a mistake in your substitution. Could you write exactly what you did? Mar 6, 2017 at 6:59
• y=|9-x²| =>y²=(9-x²)² =>2y dy/dx =2*(-2x)(9-x²) =>dy/dx = (2x³-18x)/|9-x²| Mar 6, 2017 at 7:07
• @ShoaibAshraf That is true only when $x \neq -3,3$. The denominator becomes zero at those points. Mar 6, 2017 at 7:09
• Yes understood. Thanks.. Mar 6, 2017 at 7:10

You found the points where $f(x)=0$, not where the derivative is zero.

Rolle’s theorem. If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, with $f(a)=f(b)$, then there exists $c\in(a,b)$ such that $f(c)=0$.

By the way, such a point $c$ can be chosen to be either a point of absolute minimum or a point of absolute maximum for $f$ in $(a,b)$.

The given function satisfies the hypotheses: over $(-3,3)$ we have $f(x)=9-x^2$ which is clearly differentiable. The function is continuous over $[-3,3]$ by continuity of the absolute value.

In this case there is just one point where the derivative vanishes, namely $x=0$.

However this is not general: the function $\sin x$ over $[0,2\pi]$ has the derivative that vanishes at two points. Over the interval $[0,n\pi]$ the derivative vanishes at $n$ points ($n$ positive integer).

The function $\sigma$ defined by $$\sigma(x)=\begin{cases} 0 & \text{if x=0} \\[6px] x\sin\dfrac{\pi}{x} & \text{if x\ne0} \end{cases}$$ satisfies the hypotheses of Rolle’s theorem over $[0,1]$; there are infinitely many points in $(0,1)$ where the derivative vanishes.