Chapter 4.1, Problem 36E

To determine

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

Answer to Problem 36E

Minimum occur at $x=-1,1$ and extreme value is $y=-3$ .

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

Explanation of Solution

Given information :

  $y=x^{2/3}\left(x^2-4\right)$

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 $y=x^{2/3}\left(x^2-4\right)$ with respect to $x$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x^{2/3}\left(x^2-4\right)\right)\\ =x^{2/3}\cdot \frac{d}{dx}\left(x^2-4\right)+\left(x^2-4\right)\cdot \frac{d}{dx}\left(x^{2/3}\right)\\ =x^{2/3}\cdot \left(2x\right)+\left(x^2-4\right)\cdot \left(\frac{2}{3}{x}^{-1/3}\right)\\ =2x^{5/3}+\frac{2\left(x^2-4\right)}{3x^{1/3}}\\ =\frac{6x^2+2\left(x^2-4\right)}{3x^{1/3}}\\ =\frac{8\left(x^2-1\right)}{3x^{1/3}}\end{array}$

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

  $\begin{array}{c}\frac{8\left(x^2-1\right)}{3x^{1/3}}=0\\ x^2-1=0\left[x^{1/3}\ne 0\right]\\ x=-1,1\end{array}$

The critical points are $x=-1,0,1$ .

Now plug point $x=-1$ in $y=x^{2/3}\left(x^2-4\right)$ .

  $\begin{array}{c}y={\left(-1\right)}^{2/3}\left({\left(-1\right)}^2-4\right)\\ ={\left(-1\right)}^{2/3}\left(1-4\right)\\ =-3\end{array}$

Now plug point $x=0$ in $y=x^{2/3}\left(x^2-4\right)$ .

  $\begin{array}{c}y=0^{2/3}\left(0^2-4\right)\\ =0\end{array}$

Now plug point $x=1$ in $y=x^{2/3}\left(x^2-4\right)$ .

  $\begin{array}{c}y={\left(1\right)}^{2/3}\left(1^2-4\right)\\ =\left(1-4\right)\\ =-3\end{array}$

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

As $x$ moves away from $-1$ on either side, the value of $y$ increases.

Thus, minimum occur at $x=-1$ .

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

Thus, maximum occur at $x=0$ .

As $x$ moves away from $1$ on either side, the value of $y$ increases.

Thus, minimum occur at $x=1$ .