Where is Rolle's Theorem true?

Question

for the function f(x)=x^3-2x^2-3x, find the values where Rolle's Theorem is true.

Answer 1

Rolle's theorem states If a real-valued function f(x) is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one c in the open interval (a, b) such that f'(c)=0.

 

Our function is f(x)=x³-2x²-3x.

(This function has 3 real zeros: x =-1, x =0, x =3.Do you know how to find?!)

Take derivative: f'(x)=3x²-4x-3

Lets solve equation

3x²-4x-3=0,

x₁=(1/3)(2 -√13), x₂=(1/3)(2+√13). These two points are those, where f'(x)=0. (Roughly speaking, this the result of Rolle's Theorem.)

Now we can take two intervals [a, b] and [c, d],  such that x₁∈[a, b] f(a)=f(b) ,  x₂∈[c, d] and  f(c)=f(d) 

Now question is how can we find (choose)  a, b, c and d? The good candidates are zeros of f(x). So the intervals are [-1, 0] and [0, 3]

It is clear that f(-1)=f(0)=f(3)=0. 

These two intervals are not  the only intervals. For instance, you can increase (or decrease) each of these intervals such that the condition in Rolle's theorem remains in  power, that is f(a)=f(b).

I hope this answer will help you.