Question

Let, $f\left(x\right)=x+\frac{1}{x}$ then prove that $f^{\prime }\left(1\right)=0=f(-1)$.

Answer 1

Given, $f\left(x\right)=x+\frac{1}{x}$.

Now $f^{\prime }\left(x\right)=1-\frac{1}{x^2}$

Now $f(-1)=1-1=0$ and $f(-1)=1-1=0$.

So $f^{\prime }(-1)=0=f\left(1\right)$.