Q. 40 Use the second-derivative test t... [FREE SOLUTION]

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Chapter 3: Q. 40 (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.)
$f (x) = \cos (\sin^{- 1} x)$

Short Answer

Expert verified

The function has a local maximum at $x = 1 .$

Step by step solution

Step 1. Given information.

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

Step 2. Critical points.

On differentiating, we get,

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

Therefore, the critical point is $x = 0, w h e r e f' (x) = 0$ .

Step 3. Second-derivative test.

On differentiating again, we get,

$f^{''} (x) = \frac{d}{d x} (- \frac{x}{\sqrt{1 - x^{2}}}) = - \frac{(\frac{d}{d x} x) \sqrt{1 - x^{2}} - x (\frac{d}{d x} \sqrt{1 - x^{2}})}{{(\sqrt{1 - x^{2}})}^{2}} = - \frac{\sqrt{1 - x^{2}} - x (\frac{1}{2 \sqrt{1 - x^{2}}} \frac{d}{d x} (1 - x^{2}))}{{(\sqrt{1 - x^{2}})}^{2}} = - \frac{\sqrt{1 - x^{2}} - x (\frac{- 2 x}{2 \sqrt{1 - x^{2}}})}{{(\sqrt{1 - x^{2}})}^{2}} = - \frac{\sqrt{1 - x^{2}}}{{(\sqrt{1 - x^{2}})}^{2}} - \frac{\frac{x^{2}}{\sqrt{1 - x^{2}}}}{{(\sqrt{1 - x^{2}})}^{2}} f^{''} (0) = - \frac{1}{\sqrt{1}} - \frac{0^{2}}{(\sqrt{1})^{\frac{3}{2}}} = - 1 < 0 T h e r e f o r e, i t s h o w s a l o c a l m a x i m a .$

Step 4. Verification.

The graph of the function is ,

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Short Answer

Step by step solution

Step 1. Given information.

Step 2. Critical points.

Step 3. Second-derivative test.

Step 4. Verification.

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