Q. 38 - Calculus

Question

Use the second-derivative test to determine the local extrema of each function $f$ in Exercises 29–40. 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 39–50 of Section 3.2.) 

$f\left(x\right)=si{n}^{-1}{x}^2$.

Step-by-Step Solution

Answer

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

Step 1. Given Information.

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

Step 2. Critical points.

On differentiating the given function, we get,

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{d}{dx}{\sin }^{-1}{x}^2 \\ =\frac{1}{\sqrt{1-{\left(x^2\right)}^2}}\frac{d}{dx}{x}^2 \\ =\frac{2x}{\sqrt{1-x^4}} \end{gathered}$$

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

Step 3. Second-Derivative Test.

On differentiating again, we get,

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\frac{2x}{\sqrt{1-x^4}} \\ =\frac{\left(2\frac{d}{dx}x\right)\sqrt{1-x^4}-2x\left(\frac{d}{dx}\sqrt{1-x^4}\right)}{{\left(\sqrt{1-x^4}\right)}^2} \\ =\frac{2\sqrt{1-x^4}-2x\left(\frac{1}{2\sqrt{1-x^4}}\frac{d}{dx}\left(1-x^4\right)\right)}{{\left(\sqrt{1-x^4}\right)}^2} \\ =\frac{2\sqrt{1-x^4}+\frac{4x^4}{\sqrt{1-x^4}}}{{\left(\sqrt{1-x^4}\right)}^2} \\ =\frac{2x^4+2}{{\left(1-x^4\right)}^{\frac{3}{2}}} \\ {f}^{\text{'}\text{'}}\left(0\right)=\frac{2}{(1{)}^{\frac{3}{2}}} \\ =2>0\{Local Minimum\} \end{gathered}$$

Step 4. Verification.

The graph of the function is ,

which has a local extrema at $x=0.$