Question

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

Answer 1

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

For the function

`f(x) = x^3/3-3x`

we see that

• `f(x)` is continuous and differentiable
• `f(-3)=0= f(3)`

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

`f^'(x) = 0`

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

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

Note that Rolle's theorem says that there is a `x in (a,b)` where `f^'(x)` 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]}