Q. 39 - 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.) 

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

Step-by-Step Solution

Answer

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

Step 1. Given Information.

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

Step 2. Critical points.

On differentiating the given function,

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

Therefore, the critical points are $x=0.$

Step 3. Second-derivative test.

Again differentiating the function, we get,

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}x\sin x{\cos }^{-2}x \\ =\left(\frac{d}{dx}x\right)\sin x{\cos }^{-2}x+x\left(\frac{d}{dx}\sin x\right){\cos }^{-2}x+x\sin x\left(\frac{d}{dx}{\cos }^{-2}x\right) \\ =\sin x{\cos }^{-2}x+x\cos x{\cos }^{-2}x+x\sin x\left(\frac{d}{dx}{\cos }^{-2}x\right) \\ =\frac{2x{\sin }^{2}x}{{\cos }^{3}x}+\frac{\sin x}{{\cos }^{2}x}+\frac{x}{\cos x} \\ {f}^{\text{'}\text{'}}\left(0\right)=\frac{2\left(0\right){\sin }^2\left(0\right)}{{\cos }^3\left(0\right)}+\frac{\sin \left(0\right)}{{\cos }^2\left(0\right)}+\frac{\left(0\right)}{\cos \left(0\right)}=0 \\ This shows that the function will have a local maximum at this point. \end{gathered}$$

Step 4. Verification.

The graph of the function is,