Q. 50 - 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.

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

Step-by-Step Solution

Answer

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

Step 1 . Given information

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

Step 2 . Consider the function,

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

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

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

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

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

Thus, the critical point is, $x=0$.

Step 3 . Now calculate the second derivative.

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

So, $f^{\text{'}\text{'}}\left(0\right)=-1<0$.

Now, 

$$\begin{gathered} f\left(0\right)=\cos \left({\sin }^{-1}0\right) \\ =\cos 0 \left[\because {\sin }^{-1}0=0\right] \\ =1 \end{gathered}$$

Therefore, the local maxima at $x=0$.

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