Chapter 4.1, Problem 35E

To determine

To find : extreme values of the function and where they occur.

Answer to Problem 35E

Maximum occur at $x=-\frac{4}{5}$ and extreme value is $y=\frac{6}{5}{\left(-\frac{4}{5}\right)}^{2/3}$ .

Minimum occur at $x=0$ and extreme value is $y=0$ .

Explanation of Solution

Given information :

  $y=x^{2/3}\left(x+2\right)$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

Product rule of derivative $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ .

The domain of the function is $ℝ$ .

Differentiate $y=x^{2/3}\left(x+2\right)$ with respect to $x$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x^{2/3}\left(x+2\right)\right)\\ =x^{2/3}\cdot \frac{d}{dx}\left(x+2\right)+\left(x+2\right)\cdot \frac{d}{dx}\left(x^{2/3}\right)\\ =x^{2/3}\cdot \left(1\right)+\left(x+2\right)\cdot \left(\frac{2}{3}{x}^{-1/3}\right)\\ =x^{2/3}+\frac{2\left(x+2\right)}{3x^{1/3}}\\ =\frac{3x+2\left(x+2\right)}{3x^{1/3}}\\ =\frac{5x+4}{3x^{1/3}}\end{array}$

Take $\frac{dy}{dx}=0$ for critical points,

  $\begin{array}{c}\frac{5x+4}{3x^{1/3}}=0\\ 5x+4=0\left[x^{1/3}\ne 0\right]\\ x=-\frac{4}{5}\end{array}$

The critical points are $x=-\frac{4}{5},0$ .

Now plug point $x=-\frac{4}{5}$ in $y=x^{2/3}\left(x+2\right)$ .

  $\begin{array}{c}y={\left(-\frac{4}{5}\right)}^{2/3}\left(-\frac{4}{5}+2\right)\\ ={\left(-\frac{4}{5}\right)}^{2/3}\left(\frac{-4+10}{5}\right)\\ =\frac{6}{5}{\left(-\frac{4}{5}\right)}^{2/3}\end{array}$

Now plug point $x=0$ in $y=x^{2/3}\left(x+2\right)$ .

  $\begin{array}{c}y=0^{2/3}\left(0+2\right)\\ =0\end{array}$

Examine whether the function is either maxima or minima. Check the surrounding values of critical point.

As $x$ moves away from $-\frac{4}{5}$ on either side, the value of $y$ decreases.

Thus, maximum occur at $x=-\frac{4}{5}$ .

As $x$ moves away from $0$ on either side, the value of $y$ increases.

Thus, minimum occur at $x=0$ .