Q. 49 - 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.

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

Step-by-Step Solution

Answer

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

Step 1 . Given information

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

Step 2 . Consider the function,

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

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

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

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, 

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

Therefore, the local maxima at $x=\frac{\mathrm{\pi }}{2}$.

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