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

Question

Paisley Johnson · Accepted Answer

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

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]}

Emma Aldridge · Answer

undefined To use Rolle's Theorem for $ f(x) = \frac{x^3}{3} - 3x $ on the interval $[-3, 3]$, you need to check three conditions:

1. Verify that the function $ f(x) $ is continuous on the closed interval $[-3, 3]$.
2. Confirm that $ f(x) $ is differentiable on the open interval $(-3, 3)$.
3. Ensure that $ f(-3) = f(3) $.

If all three conditions are met, then according to Rolle's Theorem, there exists at least one value $ c $ in the open interval $(-3, 3)$ such that $ f'(c) = 0 $.