Chapter 13.6, Problem 44PPSE

Solution

To write: The equation for the graph of the tangent function shown below.

  

The tangent function is written as $y=\frac{1}{2}\tan \left(\frac{1}{2}x\right)$ .

Given:

The graph of the tangent function is shown below.

  

Concept used:

The tangent curve is of the form $y=a\tan \left(bx\right)$ where $a\ne 0$ and $b>0$ has period $\left(T\right)=\frac{\pi }{b}$ , the number of cycles in the interval from $-\frac{\pi }{2b}\text{ to }\frac{\pi }{2b}$ is $1$ and two vertical asymptotes $-\frac{\pi }{2b},\frac{\pi }{2b}$ at each end of the cycle.

Calculation:

From the graph of the function, it can be interpreted that the one cycle of the tangent curve is from $-\pi \text{ to }\pi$ where the vertical asymptotes of the function exist. Therefore, the period of the function is difference of one cycle length which is $2\pi$ . Therefore, evaluate the number of cycles.

  $\begin{array}{c}\text{Period}\left(T\right)=\frac{\pi }{b}\\ 2\pi =\frac{\pi }{b}\mathrm{....}\left(\because T=2\pi \right)\\ \text{or, }b=\frac{1}{2}\end{array}$

And the point $\left(\frac{\pi }{2},\frac{1}{2}\right)$ lies in the graph of function. Therefore, satisfy the point $\left(\frac{\pi }{2},\frac{1}{2}\right)$ in the tangent function $y=a\tan \left(bx\right)$ gives,

  $\begin{array}{c}y=a\tan \left(bx\right)\\ \frac{1}{2}=a\tan \left(\frac{1}{2}\cdot \left(\frac{\pi }{2}\right)\right)\mathrm{....}\left(\begin{array}{l}\text{Satisfy point }\left(\frac{\pi }{2},\frac{1}{2}\right)\\ \text{and substitute }b=\frac{1}{2}\end{array}\right)\\ \text{or, }a\tan \left(\frac{\pi }{4}\right)=\frac{1}{2}\\ a=\frac{1}{2}\end{array}$

Hence, the tangent function is written as $y=\frac{1}{2}\tan \left(\frac{1}{2}x\right)$ by substituting $\left(a\right)=\frac{1}{2}\text{ and }\left(b\right)=\frac{1}{2}$ in the function $y=a\tan \left(bx\right)$ .

Conclusion:

The tangent function is written as $y=\frac{1}{2}\tan \left(\frac{1}{2}x\right)$ .