Q. 48 - Calculus

Question

Use the first-derivative test to determine the local extrema of each function $f$ in Exercises $39-50$. Then verify your algebraic answers with graphs from a calculator or graphing utility.

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

Step-by-Step Solution

Answer

The function $f\left(x\right)={\sin }^{-1}{x}^2$ has local minima at $x=0$. The following graph verifies the algebraic result graphically:

Step 1 . Given information

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

Step 2 . Consider the function,

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

Find the derivative of the function using the chain rule as follows:

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

The derivative is defined and continuous and only on the interval $\left(-1,-1\right)$, so the critical points of $f$ are the points where $f^{\text{'}}\left(x\right)=0$, that is,

$$\begin{gathered} \frac{2x}{\sqrt{1-x^4}}=0 \\ 2x=0 \\ x=0 \end{gathered}$$

Thus, the critical point is at $x=0$.

Step 3 . Now calculate the second derivative.

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left(\frac{2x}{\sqrt{1-x^4}}\right) \\ =\frac{\sqrt{1-x^4}\left(2\right)-2x{\displaystyle \frac{1}{2\sqrt{1-x^4}}}\left(-4x^3\right)}{1-x^4} \\ =\frac{2\left(x^4+1\right)}{{\left(-x^4+1\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \end{gathered}$$

So, $f^{\text{'}\text{'}}\left(0\right)=2>0$.

Now evaluate the function at the critical points.

$$\begin{gathered} f\left(0\right)={\sin }^{-1}{\left(0\right)}^2 \\ =0 \end{gathered}$$

Therefore, the local minima at $x=0$.

Step 4 . The following graph verifies the algebraic result graphically: