Q. 38 - 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.

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}$

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:

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{sinx}}$

Step 2. Finding the derivative of the function:

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

Step 3. Finding increasing and decreasing intervals:

Intervals of the given function :

$$\begin{gathered} f^{\text{'}}\left(x\right) has x=0 \\ {f}^{\text{'}}\left(0\right)=\cos (2.0) \\ =\cos 0 \\ =1>0 \\ {f}^{\text{'}}\left(\frac{\pi }{2}\right)=\cos (2.\frac{\pi }{2}) \\ =\cos \pi \\ =-1<0 \end{gathered}$$

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 :