Question

For the function $f\left(x\right)=\frac{1}{x^2}$, there exists value of x in the interval [-2,2] where $f^{\prime }\left(x\right)=0$

• A. exactly one
• B. at least one
• C. no

Answer 1

The correct option is C no

According to Rolle's mean value theorem, if f(x) is continuous in the interval $[\alpha ,\beta ]$ and differentiable in the interval $(\alpha ,\beta )$, and if $f\left(\alpha \right)=f\left(\beta \right)$, then, there exists a value c such that $\alpha <c<\beta$ and $f^{\prime }\left(c\right)=0$.

Here, $f\left(x\right)=\frac{1}{x^2}$ -This function is not continuous at x = 0, as f(0) is not defined. Hence, there exists no value of x in the interval [-2,2] such that f'(x)=0