Chapter 4.4, Problem 17E

To determine

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

Answer to Problem 17E

Increasing on $[-4.91,-4.51),(0,1.77),(4.91,8.06),(10.79,14.34),(17.08,20)$

Decreasing on $(-4.51,-4.10),(1.77,4.10),(8.06,10.79),(14.34,17.08)$

Local minima: $f(-4.51)\approx 0.62,f(1.77)\approx 2.58,f(8.06)\approx 3.60,f(14.34)\approx 4.39$

Local maxima

  $\begin{array}{l}f(10.79)\approx 2.43\\ f(17.08)\approx 3.49\end{array}$

Concave down on $(-4.91,-4.10),(0,4.10),(4.91,9.60),(12.25,15.81),(18.6,20)$

Concave up on $(9.60,12.25),(15.81,18.6)$

Inflection point is at

  $\begin{array}{l}f(9.60)\approx 2.9\\ f(12.25)\approx 3.27\\ f(15.81)\approx 3.91\\ f(18.65)\approx 4.2\end{array}$

Explanation of Solution

Given:Use computer algebra system to graph $f$ . Find $f\text{'}andf\text{'}\text{'}$ and use their graph to estimate the interval of increase and decrease and concavity of $f$ .

  $f(x)=\sqrt{x+5\sin x},x\le 20$

Given: $f(x)=\sqrt{x+5\sin x},x\le 20$

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

Then

  $\begin{array}{l}f\text{'}(x)=-\frac{1+5\cos x}{2\sqrt{x+5\sin x}}\\ f\text{'}\text{'}(x)=\frac{1+10\cos x+25{\cos }^{2}x+10\sin x+50{\sin }^{2}x}{4\sqrt{{(x+5\sin x)}^3}}\end{array}$

The graph for $f$

  

The graph for $f\text{'}$

  

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

  

From the graph it is conclude that

Increasing on $[-4.91,-4.51),(0,1.77),(4.91,8.06),(10.79,14.34),(17.08,20)$

Decreasing on $(-4.51,-4.10),(1.77,4.10),(8.06,10.79),(14.34,17.08)$

Local minima

  $\begin{array}{l}f(-4.51)\approx 0.62\\ f(1.77)\approx 2.58\\ f(8.06)\approx 3.60\\ f(14.34)\approx 4.39\end{array}$

Local maxima

  $\begin{array}{l}f(10.79)\approx 2.43\\ f(17.08)\approx 3.49\end{array}$

Concave down on $(-4.91,-4.10),(0,4.10),(4.91,9.60),(12.25,15.81),(18.6,20)$

Concave up on $(9.60,12.25),(15.81,18.6)$

Inflection point is at

  $\begin{array}{l}f(9.60)\approx 2.9\\ f(12.25)\approx 3.27\\ f(15.81)\approx 3.91\\ f(18.65)\approx 4.2\end{array}$