Q. 38 Use the second-derivative test t... [FREE SOLUTION]

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Chapter 3: Q. 38 (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.)
$f (x) = s i n^{- 1} x^{2}$ .

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information.

The given function is $f (x) = \sin^{- 1} x^{2} .$

Step 2. Critical points.

On differentiating the given function, we get,

$f^{'} (x) = \frac{d}{d x} \sin^{- 1} x^{2} = \frac{1}{\sqrt{1 - {(x^{2})}^{2}}} \frac{d}{d x} x^{2} = \frac{2 x}{\sqrt{1 - x^{4}}}$

The derivative is zero at $x = 0,$ therefore, $x = 0$ is the critical point.

Step 3. Second-Derivative Test.

On differentiating again, we get,

$f^{''} (x) = \frac{d}{d x} \frac{2 x}{\sqrt{1 - x^{4}}} = \frac{(2 \frac{d}{d x} x) \sqrt{1 - x^{4}} - 2 x (\frac{d}{d x} \sqrt{1 - x^{4}})}{{(\sqrt{1 - x^{4}})}^{2}} = \frac{2 \sqrt{1 - x^{4}} - 2 x (\frac{1}{2 \sqrt{1 - x^{4}}} \frac{d}{d x} (1 - x^{4}))}{{(\sqrt{1 - x^{4}})}^{2}} = \frac{2 \sqrt{1 - x^{4}} + \frac{4 x^{4}}{\sqrt{1 - x^{4}}}}{{(\sqrt{1 - x^{4}})}^{2}} = \frac{2 x^{4} + 2}{{(1 - x^{4})}^{\frac{3}{2}}} f^{''} (0) = \frac{2}{(1)^{\frac{3}{2}}} = 2 > 0 {L o c a l M i n i m u m}$