Chapter 7: Q. 77 (page 458)

In Problems 67–78,
$f (x) = \sin x, - \frac{π}{2} \leq x \leq \frac{π}{2}, g (x) = \cos x, 0 \leq x \leq π and h (x) = tanx, - \frac{π}{2} < x < \frac{π}{2}$
Find the exact value of each composite function.
$h (g^{- 1} (- \frac{1}{4}))$

Short Answer

Expert verified

The required exact value of given expression is $- \sqrt{15}$ .

Step by step solution

Step 1. Given information.

The composite function $h (g^{- 1} (- \frac{1}{4}))$ .

Step 2. Here, let g-1-14=x

So,

$g (x) = - \frac{1}{4} \cos x = - \frac{1}{4} S i n c e, \cos x < 0, s o \frac{π}{2} \leq x \leq π Now, sinx = \sqrt{1 - \cos^{2} x} = \sqrt{1 - {(- \frac{1}{4})}^{2}} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$

Step 3. Now, find hg-1-14.

h (g^{- 1} (- \frac{1}{4})) = h (x) = \tan x = \frac{\sin x}{\cos x} = \frac{\sqrt{15} / 4}{- 1 / 4} = - \sqrt{15}

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In Problems 67–78,
$f (x) = \sin x, - \frac{π}{2} \leq x \leq \frac{π}{2}, g (x) = \cos x, 0 \leq x \leq π and h (x) = tanx, - \frac{π}{2} < x < \frac{π}{2}$
Find the exact value of each composite function.
$h (g^{- 1} (- \frac{1}{4}))$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Here, let g-1-14=x

Step 3. Now, find hg-1-14.

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In Problems 67–78, fx=sinx,-π2≤x≤π2,gx=cosx,0≤x≤πandhx=tanx,-π2<x<π2Find the exact value of each composite function.hg-1-14

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Here, let g-1-14=x

Step 3. Now, find hg-1-14.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

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Probability and Statistics

Study anywhere. Anytime. Across all devices.

In Problems 67–78,
$f (x) = \sin x, - \frac{π}{2} \leq x \leq \frac{π}{2}, g (x) = \cos x, 0 \leq x \leq π and h (x) = tanx, - \frac{π}{2} < x < \frac{π}{2}$
Find the exact value of each composite function.
$h (g^{- 1} (- \frac{1}{4}))$