Q. 40 - 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.)

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

Step-by-Step Solution

Answer

The function has a local maximum at $x=1.$

Step 1. Given information.

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

Step 2. Critical points.

On differentiating, we get,

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

Therefore, the critical point is $x=0, where f\text{'}\left(x\right)=0$.

Step 3. Second-derivative test.

On differentiating again, we get,

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\left(-\frac{x}{\sqrt{1-x^2}}\right) \\ =-\frac{\left(\frac{d}{dx}x\right)\sqrt{1-x^2}-x\left(\frac{d}{dx}\sqrt{1-x^2}\right)}{{\left(\sqrt{1-x^2}\right)}^2} \\ =-\frac{\sqrt{1-x^2}-x\left(\frac{1}{2\sqrt{1-x^2}}\frac{d}{dx}\left(1-x^2\right)\right)}{{\left(\sqrt{1-x^2}\right)}^2} \\ =-\frac{\sqrt{1-x^2}-x\left(\frac{-2x}{2\sqrt{1-x^2}}\right)}{{\left(\sqrt{1-x^2}\right)}^2} \\ =-\frac{\sqrt{1-x^2}}{{\left(\sqrt{1-x^2}\right)}^2}-\frac{\frac{x^2}{\sqrt{1-x^2}}}{{\left(\sqrt{1-x^2}\right)}^2} \\ {f}^{\text{'}\text{'}}\left(0\right)=-\frac{1}{\sqrt{1}}-\frac{0^2}{(\sqrt{1}{)}^{\frac{3}{2}}} \\ =-1<0 \\ Therefore, \\ it shows a local maxima. \end{gathered}$$

Step 4. Verification.

The graph of the function is ,