Answer
a. $16$.
b. $-1$.
Work Step by Step
a. Given the function $f(x)=x^2+\frac{a}{x}$, we have $f'(x)=2x-\frac{a}{x^2}$ and $f''(x)=2+\frac{2a}{x^3}$. For the function to have a local minimum at $x=2$, we have $f'(2)=0$ which gives $4-\frac{a}{4}=0$ or $a=16$. Check $f''(2)=2+\frac{32}{2^3}=6\gt0$ which confirms the value is a local minimum.
b. For the function to have a point of inflection at $x=1$, let $f''(1)=0$, we have $2+\frac{2a}{1^3}=0$ or $a=-1$. Check the signs of $f''$ across $x=1$ at $a=-1$ to get: $..(-)..(1)..(+)..$ which confirms the inflection point.
