Q. 39 Use the second-derivative test t... [FREE SOLUTION]

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Chapter 3: Q. 39 (page 275)

Use the second-derivative test to determine the local extrema of each function $f$ in Exercises 29鈥�40. If the second-derivative test fails, you may use the first-derivative test. Then verify your algebraic answers with graphs from a calculator or graphing utility. (Note: These are the same functions that you examined with the first-derivative test in Exercises 39鈥�50 of Section 3.2.)
$f (x) = \sin (\cos^{- 1} x)$

Short Answer

Expert verified

The function has a local maximum at $x = 0 .$

Step by step solution

Step 1. Given Information.

The given function is $f (x) = \sin (\cos^{- 1} x)$ .

Step 2. Critical points.

On differentiating the given function,

$f^{'} (x) = \frac{d}{d x} \sin (\cos^{- 1} x) = \cos (\cos^{- 1} x) \frac{d}{d x} (\cos x)^{- 1} = x (- (\cos x)^{- 2}) \frac{d}{d x} (\cos x) [\cos (\cos^{- 1} x) = x] = x \sin x \cos^{- 2} x$

Therefore, the critical points are $x = 0 .$

Step 3. Second-derivative test.

Again differentiating the function, we get,

$f^{''} (x) = \frac{d}{d x} x \sin x \cos^{- 2} x = (\frac{d}{d x} x) \sin x \cos^{- 2} x + x (\frac{d}{d x} \sin x) \cos^{- 2} x + x \sin x (\frac{d}{d x} \cos^{- 2} x) = \sin x \cos^{- 2} x + x \cos x \cos^{- 2} x + x \sin x (\frac{d}{d x} \cos^{- 2} x) = \frac{2 x \sin^{2} x}{\cos^{3} x} + \frac{\sin x}{\cos^{2} x} + \frac{x}{\cos x} f^{''} (0) = \frac{2 (0) \sin^{2} (0)}{\cos^{3} (0)} + \frac{\sin (0)}{\cos^{2} (0)} + \frac{(0)}{\cos (0)} = 0 T h i s s h o w s t h a t t h e f u n c t i o n w i l l h a v e a l o c a l m a x i m u m a t t h i s p o i n t .$