Chapter 4.4, Problem 44E

(a)

To determine

To explain: why only consider the values of $x$ for the function $f\left(x\right)=3+4\cos x+\cos 2x$ that lies in the interval $\left[0,2\pi \right]$ .

Answer to Problem 44E

Since the function $f\left(x\right)=3+4\cos x+\cos 2x$ contains the $\cos x$ term, only those values of x would be considered that lies in the interval $\left[0,2\pi \right]$ .

Explanation of Solution

Given information: The function $f\left(x\right)=3+4\cos x+\cos 2x$ is given.

It is known that the functions $\sin x$ and $\cos x$ are periodic function in the period or in the interval $\left[0,2\pi \right]$ .

Since the function $f\left(x\right)=3+4\cos x+\cos 2x$ contains the $\cos x$ term, only those values of x would be considered that lies in the interval $\left[0,2\pi \right]$ .

(b)

To determine

To identify whether the given function is ever negative.

Answer to Problem 44E

No

Explanation of Solution

Given information : The function $f\left(x\right)=3+4\cos x+\cos 2x$ is given.

Differentiate the function $f\left(x\right)=3+4\cos x+\cos 2x$ with respect to x on both sides.

  $\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(3+4\cos x+\cos 2x\right)\\ f^{\prime }\left(x\right)=\frac{d}{dx}\left(3\right)+4\frac{d}{dx}\left(\cos x\right)+\frac{d}{dx}\left(\cos 2x\right)\\ =0+4\left(-\sin x\right)-2\sin 2x\\ =-4\sin x-2\sin 2x\end{array}$

Now, equate the obtained derivative to zero to get the critical point of the given function where the function might have minima or maxima.

  $\begin{array}{c}-4\sin x-2\sin 2x=0\\ -4\sin \left(x\right)-4\cos \left(x\right)\sin \left(x\right)=0\\ -4\sin \left(x\right)\left(\cos \left(x\right)+1\right)=0\end{array}$

Equate each of the factors to zero and solve for x using the fact that the given interval is $\left[0,2\pi \right]$ .

  $\begin{array}{c}-4\sin x=0\\ \sin x=0\\ \cos x+1=0\\ \cos x=-1\\ x=0,\pi ,2\pi \end{array}$

Determine the second derivative of the given function by differentiating the obtained first derivative with respect to x on both sides.

  $\begin{array}{c}\frac{d}{dx}\left(f^{\prime }\left(x\right)\right)=\frac{d}{dx}\left(-4\sin x-2\sin 2x\right)\\ f^{″}\left(x\right)=-4\frac{d}{dx}\sin \left(x\right)-2\frac{d}{dx}\left(\sin 2x\right)\\ =-4\cos x-2\left(2\cos 2x\right)\\ =-4\cos x-4\cos 2x\end{array}$

Determine the value of the obtained second derivative at $x=0,\pi ,2\pi$ .

  $\begin{array}{c}{f}^{″}\left(0\right)=-4\cos 0-4\cos 0\\ =-4-4\\ =-8\\ f^{″}\left(\pi \right)=-4\cos \pi -4\cos \pi \\ =4+4\\ =8\\ f^{″}\left(2\pi \right)=-4\cos 2\pi -4\cos 2\pi \\ =-4-4\\ =-8\end{array}$

Since the second derivative is positive at $x=\pi$ and is negative at $x=0,2\pi$ the minima will occur at $x=\pi$ , while the maxima will occur at $x=0,2\pi$ .

Now, determine the value of the function at $x=0,\pi ,2\pi$ .

  $\begin{array}{c}f\left(0\right)=3+4\cos 0+\cos 0\\ =3+4+1\\ =8\\ f\left(\pi \right)=3+4\cos \pi +\cos 2\pi \\ =3-4+1\\ =0\\ f\left(2\pi \right)=3+4\cos 2\pi +\cos 4\pi \\ =3+4+1\\ =8\end{array}$

As a result, the minima will occur at $\left(\pi ,0\right)$ and the maxima will occur at $\left(0,8\right)$ , $\left(2\pi ,8\right)$

This implies that,

  $f\left(x\right)=0,8\ge 0$

As a result, at $\left(\pi ,0\right)$ , $f\left(x\right)$ has an absolute minimum and $f\left(x\right)>0$ .This implies that $f\left(x\right)$ can never be negative.