Chapter 4.4, Problem 11E

To determine

To Find:The intervals of the increase and decrease and intervals concavity of the function.

Answer to Problem 11E

Local maximum= $f(-5.57)\approx 0.018$ , $f(0.821)=-281.466$

Local minimum = $f(3)=0$

Explanation of Solution

Given:Sketch the graph by hand using asymptotes and the intercepts but not derivatives. Then use your sketch to producing graphs that display the major features of the curve. Use the graph to estimate the maximum and minimum values.

  $f(x)=\frac{(x+4){(x-3)}^2}{x^4(x-1)}$

Given: $f(x)=\frac{(x+4){(x-3)}^2}{x^4(x-1)}$

The factors of the numerator are

  $x=-4,3$

Thus the intercepts are $x=-4,3$

The Zeroes of the denominator are giving intercepts $x=0,1$

There is no $y$ intercept as $x=0$

The graph of the function is

  

Now,

The for the signs of the interval

  $\begin{array}{l}(-\infty ,-4):f(x)>0\\ (-4,0):f(x)<0\\ (0,1):f(x)<0\\ (1,3):f(x)>0\\ (3,\infty ):f(x)>0\end{array}$

Then

  $\begin{array}{l}=\frac{(x+4){(x-3)}^2}{x^4(x-1)}\\ =\frac{x^3+\mathrm{..}}{x^5-x^4}\\ thus\\ asymptotes=y=0\end{array}$

  

Local maximum= $f(-5.57)\approx 0.018$

Local minimum = $f(3)=0$

Now,

  $\begin{array}{l}\underset{x\to 0}{\lim }=\frac{(x+4){(x-3)}^2}{x^4(x-1)}=-\infty \\ \underset{x\to 1^{-}}{\lim }=\frac{(x+4){(x-3)}^2}{x^4(x-1)}=-\infty \\ \underset{x\to 1^{+}}{\lim }=\frac{(x+4){(x-3)}^2}{x^4(x-1)}=\infty \end{array}$

The function is increasing and decreasing on the interval $(0,1)$

And local maximum is $f(0.821)=-281.466$

Hence

Local maximum= $f(-5.57)\approx 0.018$ , $f(0.821)=-281.466$

Local minimum = $f(3)=0$