In Exercises 53–54, let f(x) = 2 sec x, g(x) = −2 tan x, and h(x)...

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

Catch the graph of Y equals A of B of X fa of X equals three CX and B of X equals three X minus pi divided by two with a corresponding graph here. Now to sketch this, let's first do our composition, we have Y equals A of B of X which just means we will plug in B FX into X for a FX. We have three see it of X which means our X will be replaced with three X minus pi divided by two. Now, we need to get some information from this equation. Let's first use our cosine which is the reciprocal of CC A. We can write this as three cosine three X minus pi divided by two. Now, we need some information. The base form of cosine is A. So BX minus C plus D where A is our amplitude A period and T is given by two pi divided by B C is our face shift when divided by B. And finally, our vertical shift is deep. Now we see our A as our amplitude, which means our amplitude will be three, our period will then be two pi divided by three our face shift will be our pi divided by two, divided by RB, which is three that gives us pi divided by six. Finally, our vertical shift is zero. There's no vertical shift now that we have our information, we can find points on our cosine graph the plot because we need to graph the cosine first to graph the sequin. So I found some values here and I chose to do two cycles of our period, which means we ended at four pi divided by three. So we see here we have an input X and an output Y make a little bigger. But we can see our values here. We can graph every one of these on our graph. Here, we have 00 pi divided by 63, I divided by 30 pi divided by two negative three, two pi divided by 30 five pi divided by 63 pi zero seven pi divided by six negative three and finally four pi divided by 30. Now, if we connect all our points with a curve that will be our cosign. Now looking for the second, the secret we can find from the cosine. The first thing is we can draw ascent totes for every time that our cosine graph hits zero, we will draw an ascent tode. So we have four ascent totes here to draw and we actually have our fifth one. If we count the very last point of our period, we can create herbs between everyone. These ay totes that touch our point. Other words, if we have a point at pi divided by 63, we will place the point there and make a curve upwards, approaching the ascent to, to the left and to the right. The negative one will do the same. But pointing downwards, finally, we will get our fourth curb run which gives us our second graph and that is how we draw the graph of our equation. Ok. I hope to help you solve the problem. Thank you for watching. Goodbye.

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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).