Chapter 4.4, Problem 14E

To determine

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

Answer to Problem 14E

Increasing on $x\in (-1.5,0)\cup (0,5)\cup (7.98,\infty )$

Decreasing on $x\in (-\infty ,-1.5)\cup (5,7.98)$

Concave up on $x\in (0,2)$

Concave down on $x\in (-\infty ,0)\cup (2,5)\cup (5,\infty )$

Explanation of Solution

Given: If function $f$ of exercise $12$ . Find $f\text{'}andf\text{'}\text{'}$ and use their graph to estimate the interval of increase and decrease and concavity of $f$ .

  $f(x)=\frac{{(2x+3)}^2{(x-2)}^5}{x^3{(x-5)}^2}$

Given: $f(x)=\frac{{(2x+3)}^2{(x-2)}^5}{x^3{(x-5)}^2}$

Calculating $f\text{'}andf\text{'}\text{'}$

Then

  $\begin{array}{l}f\text{'}(x)=\frac{2{(x-2)}^4(4x^4-22x^3-62x^2-120x-135)}{{(x-5)}^3{x}^4}\\ f\text{'}\text{'}(x)=\frac{2{(x-2)}^3(4x^6-56x^5+216x^4+460x^3+805x^2+1710x+5400)}{{(x-5)}^4{x}^5}\end{array}$

The graph for $f\text{'}$

  

From the graph it is conclude that

Increasing on $x\in (-1.5,0)\cup (0,5)\cup (7.98,\infty )$

Decreasing on $x\in (-\infty ,-1.5)\cup (5,7.98)$

The graph for $f\text{'}\text{'}$

  

From the graph

  $f(x)$ is concave up when $f\text{'}\text{'}(x)>0$ and vice versa

Concave up on $x\in (0,2)$

Concave down on $x\in (-\infty ,0)\cup (2,5)\cup (5,\infty )$