Chapter 4.4, Problem 33E

a.

To determine

To find: To find the local minimum of given function.

Answer to Problem 33E

The local minimum at $a=16$ .

Explanation of Solution

Given:

The function $f(x)=x^2+\frac{a}{x}$ .

Calculation:

  $\begin{array}{c}f(x)=x^2+\frac{a}{x}\\ f\text{'}(x)=2x-\frac{a}{x^2}\end{array}$

Critical points occur at

  $\begin{array}{c}2x-\frac{a}{x^2}=0\\ x^3=\frac{a}{2}\\ x={\left(\frac{a}{2}\right)}^{\frac{1}{3}}\end{array}$

For critical point at $x=2$

  $\begin{array}{c}{\left(\frac{a}{2}\right)}^{\frac{1}{3}}=2\\ a=2^{3}2\\ a=16\end{array}$

Also, for $x=2$ , $a=16$

  $\begin{array}{l}f\text{'}\text{'}(x)=2+\frac{2a}{x^3}\\ f\text{'}\text{'}(2)=2+\frac{2(16)}{2^3}\\ f\text{'}\text{'}(2)=2+4\\ f\text{'}\text{'}(2)=6>0\end{array}$

It gives the condition of minimum. Thus, $a=16$ there is a local minimum.

b.

To determine

To find: To find the point of inflexion for given function.

Answer to Problem 33E

The point of deflection is $x=1$ and $a=-1$ .

Explanation of Solution

Given:

The function $f(x)=x^2+\frac{a}{x}$ .

Calculation:

In order to have point of inflexion at $x=1$ it is required,

  $\begin{array}{c}f\text{'}\text{'}(x)=2+\frac{2a}{x^3}\\ f\text{'}\text{'}(1)=1+\frac{2(a)}{1^3}\\ f\text{'}\text{'}(1)=2+2a\\ a=-1\end{array}$

For $a=-1$

  $f\text{'}\text{'}(x)=2-\frac{2}{x^3}$

Here,

  $f\text{'}\text{'}(x)<0$ for $0<x<1$ and $f\text{'}\text{'}(x)>0$ for $x>1$ .

Hence, there is some concave part and some convex part.

This shows that there is a point of inflexion at $x=1$ and $a=-1$ .