1. ## Local Maximum

Prove that the function f(x)=x^2+(a/x) cannot have a local maximum for any value of a.

Any help would be greatly appreciated thanks.

2. Hello, Sophia27!

Prove that the function $\displaystyle f(x)\:=\:x^2+\frac{a}{x}$ cannot have a local maximum for any value of $\displaystyle a.$
We have: .$\displaystyle f(x)\:=\:x^2 + ax^{-1}$

Then: .$\displaystyle 2x -ax^{-2} \:=\:0$

Multiply by $\displaystyle x^2\!:\;\;2x^3 - a \:=\:0\quad\Rightarrow\quad x^3 \:=\:\tfrac{a}{2} \quad\Rightarrow\quad x \:=\:\sqrt[3]{\tfrac{a}{2}}$ .
(critical value)

Second derivative: .$\displaystyle f''(x) \;=\;2 + 2ax^{-3} \:=\:2 + \frac{2a}{x^3}$

Then: .$\displaystyle f''\!\left(\sqrt[3]{\tfrac{a}{2}}\right) \;=\;2 + \frac{2a}{\left(\sqrt[3]{\frac{a}{2}}\right)^3} \;=\;2 + \frac{2a}{\frac{a}{2}} \;=\;2 + 4 \:=\:{\color{red}+}6$

At our critical value, the second derivative is positive; the graph is concave up, $\displaystyle \cup$

We have one critical value and it is a local minimum.

Therefore, the function has no local maximum value.