Question

Graph the function. Be sure to label key points and show at least two cycles. Use the graph to determine the domain and the range of each function.

$y=-\frac{3}{2}\cos \left(\frac{\mathrm{\pi }}{4}x\right)+\frac{1}{2}$

Short answer

Graph of the function is:

Domain is $\left\{x\right|-\infty <x<\infty \}$.

Range is $\left\{y\right|-1\le y\le 2\}$.

Step-by-step solution

Step 1. Given information

We have been given a function.

We have to graph this function and determine the domain and the range of the function using the graph.

Step 2. Compare with the form y=Acos(ωx) and find amplitude and period.

On comparing, we get:

Amplitude $A=\frac{3}{2}$

Time period $T=\frac{2\mathrm{\pi }}{\omega }$

$$\begin{gathered} \omega =\frac{\mathrm{\pi }}{4} \\ T=8 \end{gathered}$$

The graph lies between $\frac{3}{2}$and $-\frac{3}{2}$.

One cycle starts from $x=0$and ends at $x=8$.

Step 3. Divide the interval into four subintervals and find the coordinates for the key points.

Five key points have x-coordinates as:

$0,2,4,6,8$

For the second cycles, x-coordinates of key points will be:

$10,12,14,16$

x and y coordinates for the first cycles are:

role="math" localid="1646397454423" $(0,-\frac{3}{2}),(2,0),(4,\frac{3}{2}),(6,0),(8,-\frac{3}{2})$

x and y coordinates for the second cycles are:

role="math" localid="1646397509167" $(10,0),(12,\frac{3}{2}),(14,0),(16,-\frac{3}{2})$

Step 4. Plot the graph

Step 5. Plot the graph for vertical shift of +12.

x and y coordinates for the first cycles will become:

$(0,-1),(2,\frac{1}{2}),(4,2),(6,\frac{1}{2}),(8,-1)$

x and y coordinates for the second cycles will become:

$(10,\frac{1}{2}),(12,2),(14,\frac{1}{2}),(16,-1)$

Step 6. Find the domain and range

The graph continues indefinitely to the left and to the right, the domain is the set of all real numbers. Therefore, the domain is $\left\{x\right|-\infty <x<\infty \}$.

The range of the function consists of all real numbers from $-1$to $2$, inclusive.

Therefore, the range is $\left\{y\right|-1\le y\le 2\}$.