Chapter 4.5, Problem 58E

a.

To determine

To describe the sequence of transformations from $f(x)=\sin x$ to $g(x)=4-\sin \left(2x+\frac{\pi }{2}\right)$ .

Answer to Problem 58E

The graph of given function $g(x)=4-\sin \left(2x+\frac{\pi }{2}\right)$ has reflection over $x-$ axis, a phase shift $\frac{\pi }{2}$ units to left, the horizontal compression of $\frac{1}{2}$ and vertical shift of $4$ units up.

Explanation of Solution

Given information:

Given functions $f(x)=\sin x$ and $g(x)=4-\sin \left(2x+\frac{\pi }{2}\right)$ .

Compare the given function $g(x)=4-\sin \left(2x+\frac{\pi }{2}\right)$ with $a\sin \left(bx-c\right)+d$ , we get

  $a=-1,b=2,c=\frac{\pi }{2}$ and $d=4$

For $a=-1$ , the graph is reflected over $x-axis$ ,

For $b=2$ , the horizontal compression of $\frac{1}{2}.$

For $c=\frac{\pi }{2}$ , the function has horizontal shift of $\frac{\pi }{2}$ units to the left.

Thus, a phase shift $\frac{\pi }{2}$ units to the right.

For $d=4$ , the function shift $4$ units up.

Therefore, the graph of given function $g(x)=4-\sin \left(2x+\frac{\pi }{2}\right)$ has reflection over $x-$ axis, a phase shift $\frac{\pi }{2}$ units to left, the horizontal compression of $\frac{1}{2}$ and vertical shift of $4$ units up.

b.

To determine

To sketch the graph of $g(x)=4-\sin \left(2x+\frac{\pi }{2}\right)$ .

Explanation of Solution

Given information:

Given function $g(x)=4-\sin \left(2x+\frac{\pi }{2}\right)$ .

Graph: The graph of $g(x)=4-\sin \left(2x+\frac{\pi }{2}\right)$ is

  

Interpretation: From above we observe that, the graph of given function $g(x)=4-\sin \left(2x+\frac{\pi }{2}\right)$ has reflection over $x-$ axis, a phase shift $\frac{\pi }{2}$ units to left, the horizontal compression of $\frac{1}{2}$ and vertical shift of $4$ units up..

c.

To determine

To write $g$ in terms of $f$ .

Answer to Problem 58E

  $g(x)=-f\left(2x+\frac{\pi }{2}\right)+4$

Explanation of Solution

Given information:

Given functions $f(x)=\sin x$ and $g(x)=4-\sin \left(2x+\frac{\pi }{2}\right)$ .

Since $f(x)=\sin x$ , replace $x\text{by}\left(2x+\frac{\pi }{2}\right)$ , we get $f\left(2x+\frac{\pi }{2}\right)=\sin \left(2x+\frac{\pi }{2}\right)$

Thus, $g(x)=-f\left(2x+\frac{\pi }{2}\right)+4$