Chapter 4.1, Problem 20E

To determine

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

Answer to Problem 20E

The extreme value of function is $2.911$ , minimum at $x=\sqrt{\frac{2}{3}}$ .

The extreme value of function is $5.089$ , maximum at $x=-\sqrt{\frac{2}{3}}$ .

Explanation of Solution

Given information :

  $y=x^3-2x+4$

Formula needed :

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

Differentiate $y=x^3-2x+4$ with respect to $x$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x^3-2x+4\right)\\ =\frac{d}{dx}\left(x^3\right)-2\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(4\right)\\ =3x^2-2\cdot 1+0\\ =3x^2-2\end{array}$

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

  $\begin{array}{c}3x^2-2=0\\ 3x^2=2\\ x^2=\frac{2}{3}\\ x=\pm \sqrt{\frac{2}{3}}\end{array}$

Now plug point $x=\sqrt{\frac{2}{3}}$ in $y=x^3-2x+4$ .

  $\begin{array}{c}y={\left(\sqrt{\frac{2}{3}}\right)}^3-2\left(\sqrt{\frac{2}{3}}\right)+4\\ =\frac{2}{3}\sqrt{\frac{2}{3}}-2\sqrt{\frac{2}{3}}+4\\ =-\frac{4}{3}\sqrt{\frac{2}{3}}+4\\ \approx 2.911\end{array}$

Next plug point $x=-\sqrt{\frac{2}{3}}$ in $y=x^3-2x+4$ .

  $\begin{array}{c}y={\left(-\sqrt{\frac{2}{3}}\right)}^3-2\left(-\sqrt{\frac{2}{3}}\right)+4\\ =-\frac{2}{3}\sqrt{\frac{2}{3}}+2\sqrt{\frac{2}{3}}+4\\ =\frac{4}{3}\sqrt{\frac{2}{3}}+4\\ \approx 5.089\end{array}$

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

As $x$ moves away from $\sqrt{\frac{2}{3}}$ on either side, the value of $y$ increases.

Thus, minimum occur at $x=\sqrt{\frac{2}{3}}$ and value of function is $2.911$ .

As $x$ moves away from $-\sqrt{\frac{2}{3}}$ on either side, the value of $y$ decreases.

Thus, maximum occur at $x=-\sqrt{\frac{2}{3}}$ and value of function is $5.089$ .