Chapter 5.4, Problem 34E

To determine

To prove that $f\left(x\right)=x^2+\left(\frac{a}{x}\right)$ cannot have a local maximum for any value of $a$ .

Answer to Problem 34E

Therefore $f$ cannot have a local maximum.

Explanation of Solution

Given information: Given equation is $f\left(x\right)=x^2+\left(\frac{a}{x}\right)$ .

Calculation:

Find local extrema:

  $\begin{array}{l}{f}^{\prime }\left(x\right)=2x-\left(a/x^2\right),\\ f^{\prime }\left(x\right)=\frac{2x^3-a}{x^2}\\ f^{\prime }\left(x\right)=0\\ 2x^3-a=0\\ x=\sqrt[3]{\frac{a}{2}}\\ {f^{\prime }}^{\prime }\left(x\right)=2+\frac{2a}{x^3}\\ =\frac{2x^3+2a}{x^3}\\ {f^{\prime }}^{\prime }\left(\sqrt[3]{\frac{a}{2}}\right)=\frac{2\frac{a}{2}+a}{\frac{a}{2}}\\ =4>0\end{array}$

Thus, therefore $f$ cannot have a local maximum.