Finding parameter values Problem:

What value of $a$ makes $$f(x)=x^2 + \frac {a}{x}$$ have (a) a minimum at $x=2$, (b) an inflection point at $x=1$?
(a) and (b) are separate and not dependent upon each other.

What I have done:
Not much. I wrote the derivative to try and find where $x=0$, but that is where I got stuck.
$$0=2x-ax^{-2} .$$
I'm not really sure what to do.
 A: (a) Take first derivative of $f(x)$, set it equal to 0, replace $x$ by 2 , and solve for $a$.
(b) Take second derivative of $f(x)$, set it equal to 0, replace $x$ by 1, and solve for $a$.
A: *

*An extrema (minimum/ maximum) is where the first derivative of the function is equal to zero: 


$$f'(x)=0$$ 
$$2x-\frac{a}{x^2}=0$$
We have $x=2$, so:$$2(2^3)-a=0$$
Therefore, $a$ would be 16.
As pointed out in comments, you should check whether the answer is correct:
$min\{f(x)=x^2+\frac{16}{x}\}=12$ and local minimum happens at $x=2$.
Looking at the plot might make it clearer:



*

*An inflection point happens where the second derivative is equal to zero as well as all other higher order derivatives , meaning if the second derivative is zero but the fourth derivative is non-zero, for example, then the point is not an inflection point. What we have:


$$f''(x)=2+\frac{2a}{x^3}=0$$
Thus, at $x=1$, $a$ would be -1.

A: $F(x)=x^2+a/x$
$F'(x)=2x-a/x^2$
let $f'(x)=0$ and put $x=2$
$a=8$
$F"(x)=2+2a/x^3$ and let $F"(x)=0$ and put $x=1$
$a =-2$
