Chapter 4.1, Problem 41E

To determine

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

Answer to Problem 41E

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

Maximum occur at $x=-1,3$ and extreme value is $y=5$ .

Explanation of Solution

Given information :

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

Formula needed :

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

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

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

For $x<1$ , $\frac{dy}{dx}=-2x-2$ and for $x>1$ , $\frac{dy}{dx}=-2x+6$ .

  $\begin{array}{c}-2x-2=0\\ x=-1\\ -2x+6=0\\ x=3\end{array}$

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

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

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{-{\left(1+h\right)}^2+6\left(1+h\right)-4-1}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{-1-h^2-2h+6+6h-5}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{-h\left(h-4\right)}{h}\\ =\underset{h\to 0^{+}}{\lim }-\left(h-4\right)\\ =4\end{array}$

And,

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{-{\left(1+h\right)}^2-2\left(1+h\right)+4-1}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{-1-h^2-2h-2-2h+3}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{h\left(-h-4\right)}{h}\\ =\underset{h\to 0^{-}}{\lim }\left(-h-4\right)\\ =-4\end{array}$

The function has no derivative at $x=1$ .

The critical points are $x=-1,1$ and $x=3$ .

As $x$ moves away from $-1$ and $3$ on either side, the value of $y$ decreases.

Thus, maximum occur at $x=-1$ and, $x=3$ . The value of function is $5$ .

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

Thus, minimum occur at $x=1$ . The value of function is $1$ .