Chapter 4.3, Problem 14E

(a)

To determine

To find:The interval on which $f$ is increasing.

Explanation of Solution

Given: The function is $f\left(x\right)=x^2\ln x$

Find the first derivative of the given function.

  $\begin{array}{c}{f}^{\prime }\left(x\right)=x^2\left(\frac{1}{x}\right)+\ln x\left(2x\right)\\ =x+2x\ln x\\ =x\left(1+2\ln x\right)\end{array}$

On solving further,

  $x=e^{-1/2}$

The function increases in the interval $\left(e^{-1/2},\infty \right)$ .

The function decreases in the interval $\left(0,e^{-1/2}\right)$

(b)

To determine

To find :The local maxima and minima.

Explanation of Solution

The function increases in the interval $\left(e^{-1/2},\infty \right)$ .

The function decreases in the interval $\left(0,e^{-1/2}\right)$

The local minimum at $\left(\frac{1}{\sqrt{e}},f\left(\frac{1}{\sqrt{e}}\right)\right)=\left(\frac{1}{\sqrt{e}},-\frac{1}{2e}\right)$ .

(c)

To determine

To find :The interval ofconcavity and points of inflection

Explanation of Solution

The graph is concave up when $f^{″}\left(x\right)>0$ and concave downward when $f^{″}\left(x\right)<0$ .

The derivative of $f^{\prime }\left(x\right)$ is positive when $f^{\prime }\left(x\right)$ is increasing.

The derivative of $f^{\prime }\left(x\right)$ is negative when $f^{\prime }\left(x\right)$ is decreasing

Calculate the second derivative.

  $\begin{array}{c}{f}^{″}\left(x\right)=2\ln x+2+1\\ =3+2\ln x\end{array}$

The second derivative is always positive, so $f$ is concave up everywhere.

The function is concave up for $\left(e^{-3/2},0\right)$ .

The function is concave down for $\left(0,e^{-3/2}\right)$ .

The point of inflection is $0$ .