Q. 37 - Calculus

Question

Use a sign chart for ${\mathit{f}}^{\mathbf{\text{'}}}$ to determine the intervals on which each function $\mathit{f}$ is increasing or decreasing. Then verify your algebraic answers with graphs from a calculator or graphing utility.

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{.}\mathit{c}\mathit{o}\mathit{s}\mathit{x}$

Step-by-Step Solution

Answer

Ans: Increasing interval $\mathbf{[}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{+}\mathbf{\pi k}\mathbf{,}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{+}\mathbf{\pi k}\mathbf{]}$

and decreasing elsewhere.

Step 1. Given information:

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{.}\mathit{c}\mathit{o}\mathit{s}\mathit{x}$

Step 2. Finding the derivative of the function:

$$\begin{gathered} f\left(x\right)=\sin x.\cos x \\ {f}^{\text{'}}\left(x\right)=\sin x(-\sin x)+\cos x\cos x \\ {f}^{\text{'}}\left(x\right)= {\cos }^{2}x-{\sin }^{2}x \\ {f}^{\text{'}}\left(x\right)=\cos 2x \\ let f^{\text{'}}\left(x\right)=0 \\ \therefore \cos 2x =0 \\ \Rightarrow 2x=\left(2k+1\right)\frac{\pi }{2} \\ \Rightarrow x=\left(2k+1\right)\frac{\pi }{4} [where k is any integer] \\ taking point x=0 \end{gathered}$$

Step 3. Finding increasing and decreasing intervals:

Intervals of the given function :
$$\begin{gathered} f^{\text{'}}\left(x\right) has x=0 \\ {f}^{\text{'}}\left(0\right)=\cos (2.0) \\ =\cos 0 \\ =1 \\ >0 \end{gathered}$$

f(x)is increasing on the interval $\mathbf{[}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{+}\mathbf{\pi k}\mathbf{,}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{+}\mathbf{\pi k}\mathbf{]}$

and decreasing elsewhere.

Step 4. Verifying algebraic answers with graphs :