Chapter 4.1, Problem 42E

To determine

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

Answer to Problem 42E

Minimum occur at $x=\frac{2\left(3+\sqrt{3}\right)}{3}$ and extreme value is $y=-2\left(16+8\sqrt{3}\right)+\frac{288+128\sqrt{3}}{9}$ .

Maximum occur at $x=-1$ and extreme value is $y=4$ .

Explanation of Solution

Given information :

  $y=\left\{\begin{array}{l}-\frac{1}{4}{x}^2-\frac{1}{2}x+\frac{15}{4},x\le 1\\ x^3-6x^2+8x,x>1\end{array}\right.$

Formula needed :

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

Differentiate $y=\left\{\begin{array}{l}-\frac{1}{4}{x}^2-\frac{1}{2}x+\frac{15}{4},x\le 1\\ x^3-6x^2+8x,x>1\end{array}\right.$ with respect to $x$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(\left\{\begin{array}{l}-\frac{1}{4}{x}^2-\frac{1}{2}x+\frac{15}{4},x\le 1\\ x^3-6x^2+8x,x>1\end{array}\right.\right)\\ =\left\{\begin{array}{l}\frac{d}{dx}\left(-\frac{1}{4}{x}^2-\frac{1}{2}x+\frac{15}{4}\right),x<1\\ \frac{d}{dx}\left(x^3-6x^2+8x\right),x>1\end{array}\right.\\ =\left\{\begin{array}{l}-\frac{1}{2}x-\frac{1}{2},x<1\\ 3x^2-12x+8,x>1\end{array}\right.\end{array}$

For $x<1$ , $\frac{dy}{dx}=-\frac{1}{2}x-\frac{1}{2}$ and for $x>1$ , $\frac{dy}{dx}=3x^2-12x+8$ .

  $\begin{array}{c}-\frac{1}{2}x-\frac{1}{2}=0\\ x=-1\\ 3x^2-12x+8=0\\ x=\frac{2\left(3+\sqrt{3}\right)}{3},\frac{2\left(3-\sqrt{3}\right)}{3}\end{array}$

Check the value of $\frac{dy}{dx}$ at $x=1$ .

Find the right and left derivative at $x=1$ .

  $\underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=-1$

And,

  $\underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=-1$

The critical points are $x=-1$ and $x=\frac{2\left(3+\sqrt{3}\right)}{3}$ .

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

Thus, maximum occur at $x=-1$ . The value of function is $4$ .

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

Thus, minimum occur at $x=\frac{2\left(3+\sqrt{3}\right)}{3}$ . The value of function is $-2\left(16+8\sqrt{3}\right)+\frac{288+128\sqrt{3}}{9}$ .