Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 223: 40

$g(x)=\dfrac{-1}{x}+x^2-1$

Here, the derivative is given as: $f'(x)=\dfrac{1}{x^2}+2x$ , thus, the general form of the function is: $f(x)=\dfrac{-1}{x}+x^2+C$ Since the graph of the given function passes through $(-1,1)$, $1+1+C=1 \implies C=-1$ Hence, $g(x)=\dfrac{-1}{x}+x^2-1$