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Chapter 4: Problem 7

Graph two periods of the given tangent function. $$y=\frac{1}{2} \tan 2 x$$

Short Answer

Expert verified
The function \(y=\frac{1}{2} \tan 2x\) has a period of \(\frac{\pi}{2}\). The function approaches but never reaches values of x at the vertical asymptotes. The function is repeated at intervals equal to the period, so two periods of this function on the graph range from 0 to \(\pi\).

Step by step solution

01

Define the period of the function

First, calculate the period of the tangent function. The period \(P\) can be calculated by dividing \(\pi\) by the absolute value of the coefficient of \(x\), in this case, 2. Therefore, the period \(P = \frac{\pi}{2}\).
02

Create a table for the function values

To graph the function, a table with values of \(x\) and corresponding values of \(y=\frac{1}{2} \tan 2x\) are needed. Pick values for \(x\) that are multiples of the period over 4 (In this case \(\frac{P}{4} = \frac{\pi}{8}\)). The typical points to consider for one period in the tangent function are 0, and \(卤\frac{P}{2}\), which are equal to 0 and \(卤\infty\) respectively. Two periods will be graphed, so the x-values will range from 0 to \(2P = \pi\).
03

Plot the values on the graph

Now, plot the points from the table created on the graph. Draw a curve through the points for positive and negative halves of the tangent function. The function will approach but never reaches the vertical asymptotes which are x = \(卤\frac{P}{2}\), \(卤\frac{3P}{2}\), ... , 卤n\(\frac{P}{2}\). Draw these asymptotes as dotted lines.

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