Chapter 10.2, Problem 7E

Solution

To match: The given graph with corresponding function.

The graph of function $y=\cos \left(\frac{x}{2}\right)+1$ is (A).

Given data: The given function is $y=\cos \left(\frac{x}{2}\right)+1$. The listed graphs are:

  

Method/Formula used:

First determine the period of the given function and some points, and match the graph.

The period $T$of a function $f\left(x\right)$ ) is defined as $f\left(x\right)=f\left(x+T\right)$ .

Calculations:

The given function is $y=\cos \left(\frac{x}{2}\right)+1$ .

Let $T$ be the period of this function, then

  $\cos \left(\frac{x}{2}\right)+1=y=\cos \left(\frac{x}{2}+T\right)+1$ … (1)

But the cosine function has period from $x=0$ to $x=2\pi$ , therefore, the function $\cos \left(\frac{x}{2}\right)+1$ has period from $\frac{x}{2}=0$ to $\frac{x}{2}=2\pi$ or from $x=0$ to $x=4\pi$ .

Thus, $4\pi$ is the period of the function $y=\cos \left(\frac{x}{2}\right)+1$ .

Substitute 0 for x in the function,

  $\begin{array}{c}y\left(0\right)=\cos \left(\frac{0}{2}\right)+1\\ =1+1\\ =2\end{array}$

Thus, the curve passes through point $\left(0,2\right)$ .

Substitute $2\pi$ for x in the function,

  $\begin{array}{c}y\left(2\pi \right)=\cos \left(\frac{2\pi }{2}\right)+1\\ =-1+1\\ =0\end{array}$

Thus, the curve passes through point $\left(2\pi ,0\right)$ .

The graph (A) has time period $4\pi$ passes through points $\left(0,2\right)$ and $\left(2\pi ,0\right)$ .

Therefore, the graph of function $y=\cos \left(\frac{x}{2}\right)+1$ is