# How to use rolles theorem for f(x)= (x^3/3)- 3x on the interval [-3,3]?

Jun 5, 2018

Rolle's theorem states that if a continuous differentiable function $f \left(x\right)$ satisfies $f \left(a\right) = f \left(b\right) = 0$, $a < b$, then there is a point $x \in \left(a , b\right)$ where ${f}^{'} \left(x\right)$ vanishes.

For the function

$f \left(x\right) = {x}^{3} / 3 - 3 x$

we see that

• $f \left(x\right)$ is continuous and differentiable
• $f \left(- 3\right) = 0 = f \left(3\right)$

Thus the conditions of Rolle's theorem are satisfied with $a = - 3 , b = 3$ and so there is a $x \in \left(- 3.3\right)$ which satisfies

${f}^{'} \left(x\right) = 0$

Since ${f}^{'} \left(x\right) = {x}^{2} - 3$ in this case, we can see that here the derivative vanishes at two points $\pm \sqrt{3}$ in the interval $\left(- 3 , 3\right)$.

(This could have been anticipated from the fact that in this case $f \left(x\right)$ is an odd, and hence ${f}^{'} \left(x\right)$ and even function, unless the value of $x$ satisfying ${f}^{'} \left(x\right) = 0$ happens to be 0, there must be another one at $- x$)

Note that Rolle's theorem says that there is a $x \in \left(a , b\right)$ where ${f}^{'} \left(x\right)$ will vanish, not that there will necessarily be only one!

This can be easily seen from the graph

graph{x^3/3-3 x [-5, 5, -5, 5]}