Question

In Exercises 53–54, let f(x) = 2 sec x, g(x) = −2 tan x, and h(x) = 2x − π/2. Graph two periods of y = (f∘h)(x).

Accepted Answer

Understand the composition of functions: The function $$ (f \circ h)(x) $$ means $$ f(h(x)) . $$Given $$ f(x) = 2 \sec x $$ and $$ h(x) = 2x - \frac{\pi}{2} $$, substitute $$ h(x) $$ into $$ f  to $$get $$ (f \circ h)(x) = 2 \sec(2x - \frac{\pi}{2}) . $$Recall the period of the secant function: The basic secant function $$ \sec x $$ has a period of $$ 2\pi . $$When the argument is modified to $$ 2x - \frac{\pi}{2} $$, the period changes according to the coefficient of $$ x . $$Calculate the new period: For a function $$ \sec(bx + c) $$, the period is $$ \frac{2\pi}{|b|} . $$Here, $$ b = 2 $$, so the period of $$ \sec(2x - \frac{\pi}{2})  is$$ $$ \frac{2\pi}{2} = \pi . $$Determine the interval for two periods: Since one period is $$ \pi $$, two periods will span $$ 2 	imes \pi = 2\pi . So$$, the graph of $$ y = 2 \sec(2x - \frac{\pi}{2}) $$ should be drawn over an interval of length $$ 2\pi  in$$ $$ x . $$Identify key points and asymptotes: To graph the function, find where the argument $$ 2x - \frac{\pi}{2} $$ equals values that cause vertical asymptotes in $$ \sec $$, such as $$ \frac{\pi}{2} + k\pi $$, and plot points accordingly to capture the shape over two periods.

In Exercises 53–54, let f(x) = 2 sec x, g(x) = −2 tan x, and h(x) = 2x − π/2. Graph two periods of y = (f∘h)(x).

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Key Concepts

Function Composition

Secant Function and Its Properties

Period of a Composite Trigonometric Function