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Chapter 4: Problem 37

Sketch the graph of the function. (Include two full periods.) $$y=\frac{1}{4} \csc \left(x+\frac{\pi}{4}\right)$$

Short Answer

Expert verified
The graph of the function \(y=\frac{1}{4} \csc \left(x+\frac{\pi}{4}\right)\) is the graph of a standard cosecant function, \(y=\csc(x)\) vertically compressed by a factor of \(\frac{1}{4}\) and shifted to left by \(\frac{\pi}{4}\) units with the same period. It has two complete periods starting from \(-\frac{\pi}{4}\) and ending at \(\frac{7\pi}{4}\) with vertical asymptotes at x values -\(\frac{\pi}{4}\), \(\frac{3\pi}{4}\), \(\frac{7\pi}{4}\).

Step by step solution

01

Identifying the period of the function

The standard period of the \(\csc(x)\) function is \(2\pi\). The given function doesn't have a coefficient with \(x\), thus, the period of this function remains \(2\pi\). Hence, the graph will start at \(-\frac{\pi}{4}\) due to the phase shift and will end at \(2\pi - \frac{\pi}{4} = \frac{7\pi}{4}\).
02

Drawing the axes and marks

Draw the x and y axes on the graph. Mark the x-axis in increments from \(-\frac{\pi}{4}\) to \(\frac{7\pi}{4}\).
03

Identifying the maximums and minimums

The csc function has no maximums and minimums, instead it has a series of vertical asymptotes at points where the sine function equals zero. The vertical asymptotes for the given function will be at x values where \(\sin(x + \frac{\pi}{4})=0\) i.e. at \(x = -\frac{\pi}{4} , \frac{3\pi}{4} , \frac{7\pi}{4}\).
04

Plotting the graph

Now plot the function \(y=\frac{1}{4} \csc (x+\frac{\pi}{4})\). Notice the function starts at \(x=-\frac{\pi}{4}\) and ends at \(x=\frac{7\pi}{4}\). Plot the vertical asymptotes at aforementioned x values. Now plot the curve in two separate parts for each period between the vertical asymptotes. Remember that the csc function is always positive in the 2nd and 4th quadrants and negative in the 1st and 3rd quadrants. Since this graph is vertically compressed by a factor of \(\frac{1}{4}\), the maximum and minimum y value will be \(\pm \frac{1}{4}\) respectively.

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