Chapter 5.2, Problem 27E

a.

To determine

To find : The local extrema of the function $f\left(x\right)=x^3-2x-2\cos x$ .

Answer to Problem 27E

The maximum is at $x=-1.126$ and the maximum value is $-0.036$ and the minimum is at 0.559 and the minimum value is $-2.639$ .

Explanation of Solution

Given information :

The given function is $f\left(x\right)=x^3-2x-2\cos x$

Calculation :

To find the local extrema first find the derivative of $f\left(x\right)$ . Therefore,

  $\begin{array}{l}f\left(x\right)=x^3-2x-2\cos x\\ f\text{'}\left(x\right)=3x^2-2-2\left(-\sin x\right)\\ f\text{'}\left(x\right)=3x^2-2+2\sin x\end{array}$

Set $f\text{'}\left(x\right)=0$ and solve for $x$ to using calculator:

  $\begin{array}{l}3x^2-2+2\sin x=0\\ x=-1.126,\text{ 0}\text{.559}\end{array}$

Substitute $x=-1.126$ in the given equation and solve:

  $\begin{array}{l}f\left(x\right)=x^3-2x-2\cos x\\ f\left(-1.126\right)={\left(-1.126\right)}^3-2\left(-1.126\right)-2\cos \left(-1.126\right)\\ f\left(-1.126\right)=-0.036\text{ Maximum}\end{array}$

Also, substitute $x=0.559$ in the given equation and solve:

  $\begin{array}{l}f\left(x\right)=x^3-2x-2\cos x\\ f\left(0.559\right)={\left(0.559\right)}^3-2\left(0.559\right)-2\cos \left(0.559\right)\\ f\left(0.559\right)=-2.639\text{ Minimum}\end{array}$

Therefore, the maximum is at $x=-1.126$ and the maximum value is $-0.036$ and the minimum is at 0.559 and the minimum value is $-2.639$ .

Hence,

The maximum is at $x=-1.126$ and the maximum value is $-0.036$ and the minimum is at 0.559 and the minimum value is $-2.639$ .

b.

To determine

To find : The interval on which the function $f\left(x\right)=x^3-2x-2\cos x$ is increasing.

Answer to Problem 27E

The function is increasing in the interval $\left(-\infty ,-1.126\right]\cup \left[0.559,\infty \right)$ .

Explanation of Solution

Given information :

The given function is $f\left(x\right)=x^3-2x-2\cos x$

Calculation :

To find the interval on which the function is increasng first find the derivative of $f\left(x\right)$ . Therefore,

  $\begin{array}{l}f\left(x\right)=x^3-2x-2\cos x\\ f\text{'}\left(x\right)=3x^2-2-2\left(-\sin x\right)\\ f\text{'}\left(x\right)=3x^2-2+2\sin x\end{array}$

Set $f\text{'}\left(x\right)=0$ and solve for $x$ to using calculator:

  $\begin{array}{l}3x^2-2+2\sin x=0\\ x=-1.126,\text{ 0}\text{.559}\end{array}$

Use sign chart to get the interval in which the function is increasing

  

Therefore, the function is increasing in the interval $\left(-\infty ,-1.126\right]\cup \left[0.559,\infty \right)$ .

Hence,

The function is increasing in the interval $\left(-\infty ,-1.126\right]\cup \left[0.559,\infty \right)$ .

c.

To determine

To find : The interval on which the function $f\left(x\right)=x^3-2x-2\cos x$ is decreasing.

Answer to Problem 27E

The interval on which the function is decreasing is $\left(-\infty ,\infty \right)-\left\{2,-2\right\}$ .

Explanation of Solution

Given information :

The given function is $f\left(x\right)=x^3-2x-2\cos x$

Calculation :

To find the interval on which the function is decreasing first find the derivative of $f\left(x\right)$ . Therefore,

  $\begin{array}{l}f\left(x\right)=x^3-2x-2\cos x\\ f\text{'}\left(x\right)=3x^2-2-2\left(-\sin x\right)\\ f\text{'}\left(x\right)=3x^2-2+2\sin x\end{array}$

Set $f\text{'}\left(x\right)=0$ and solve for $x$ to using calculator:

  $\begin{array}{l}3x^2-2+2\sin x=0\\ x=-1.126,\text{ 0}\text{.559}\end{array}$

Use sign chart to get the interval in which the function is decreasing.

  

Therefore, the function is decreasing on the interval $\left[-1.126,0.599\right]$ .

Hence,

The function is decreasing in the interval $\left[-1.126,0.599\right]$ .