Chapter 4.2, Problem 27E

(a)

To determine

To find : the local extrema.

Answer to Problem 27E

Maximum occur at $x=-1.126$ .

Minimum occur at $x=0.559$ .

Explanation of Solution

Given information :

  $f\left(x\right)=x^3-2x-2\cos x$

Formula needed :

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

Differentiate $f\left(x\right)=x^3-2x-2\cos x$ with respect to $x$ ,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(x^3-2x-2\cos x\right)\\ =\frac{d}{dx}\left(x^3\right)-\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(2\cos x\right)\\ =3x^2-2-\left(-2\sin x\right)\\ =3x^2-2+2\sin x\end{array}$

Take $f^{\prime }\left(x\right)=0$ then $3x^2-2+2\sin x=0$ .

Using online tool, $x=-1.126,0.559$ .

As $x$ move away from $-1.126$ from both ends, the value of function decreases.

Maximum occur at $x=-1.126$ .

As $x$ move away from $0.559$ from both ends, the value of function increases.

Minimum occur at $x=0.559$ .

(b)

To determine

To find : the interval in which the function is increasing.

Answer to Problem 27E

  $f\left(x\right)$ is increasing in the interval $\left(-\infty ,-1.126\right)$ and $\left(0.559,\infty \right)$ .

Explanation of Solution

Given information :

  $f\left(x\right)=x^3-2x-2\cos x$

Formula needed :

For increasing, $f^{\prime }\left(x\right)>0$

Critical points are $x=-1.126,0.559$ .

So, the intervals are $\left(-\infty ,-1.126\right),\left(-1.126,0.559\right),\left(0.559,\infty \right)$

Observe the behaviour of $f^{\prime }\left(x\right)$ ,

  $f^{\prime }\left(x\right)$ is greater than zero in the interval $\left(-\infty ,-1.126\right)$ and $\left(0.559,\infty \right)$ .

Hence, $f\left(x\right)$ is increasing in the interval $\left(-\infty ,-1.126\right)$ and $\left(0.559,\infty \right)$ .

(c)

To determine

To find : the interval in which the function is decreasing.

Answer to Problem 27E

  $f\left(x\right)$ is decreasing in the interval $\left(-1.126,0.559\right)$ .

Explanation of Solution

Given information :

  $f\left(x\right)=x^3-2x-2\cos x$

Formula needed :

For decreasing, $f^{\prime }\left(x\right)<0$

Critical points are $x=-1.126,0.559$ .

So, the intervals are $\left(-\infty ,-1.126\right),\left(-1.126,0.559\right),\left(0.559,\infty \right)$

Observe the behaviour of $f^{\prime }\left(x\right)$ ,

  $f^{\prime }\left(x\right)$ is less than zero in the interval $\left(-1.126,0.559\right)$ .

Hence, $f\left(x\right)$ is decreasing in the interval $\left(-1.126,0.559\right)$ .