Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 198: 38

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

Since, the derivative of the function is given as: $f'(x)=(\dfrac{1}{x^2}+2x)$ General form of the function will be: $f(x)=\dfrac{-1}{x}+x^2+C$ Here, the graph of the given function passes through $(-1,1)$. So, $f(x)=\dfrac{-1}{x}+x^2+C$ This implies, $1+1+C=1$ or, $C=-1$ Thus, $g(x)=\dfrac{-1}{x}+x^2-1$