Question

According to Rolle’s theorem, there will be at least one solution for f’(x) = 0 in the interval [-2,2], where $f\left(x\right)=\frac{1}{x^2}$

• A. True
• B. False

Answer 1

The correct option is B

False

Rolle’s theorem states that if we have a function which is continuous in [a,b] , differentiable in (a, b) and f(a) = f(b), then there is at least one c , between a to b such that f’(c) = 0. Here, the value of the function at x = 2 and x = -2 are equal $(=\frac{1}{2^2})$. So, can we say Rolle’s theorem is applicable here and there should be at least one solution for f’(x) = 0?

We should also check if the function is continuous in the given interval. As we can see from the graph of $\frac{1}{x^2}$, it is discontinuous at x = 0. So, we can’t apply Rolle’s theorem here.

In fact $f^{\prime }\left(x\right)=\frac{-2}{x^3}$ has no solution in the interval [-2,2]