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.

role="math" localid="1648368106886" $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\left(\frac{\pi }{2}x\right)$

Short answer

Ans: Increasing interval $\mathbf{[}\mathbf{2}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{k}\mathbf{+}\mathbf{3}\mathbf{]}$

and decreasing elsewhere.

Step-by-step solution

Step 1. Given information:

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{x}\mathbf{\right)}$

Step 2. Finding the derivative of the function:

$$\begin{gathered} f\left(x\right)=\sin \left(\frac{\pi }{2}x\right) \\ {f}^{\text{'}}\left(x\right)=\cos \left(\frac{\pi }{2}x\right).\frac{\pi }{2} \\ {f}^{\text{'}}\left(x\right)=\frac{\pi }{2}\cos \left(\frac{\pi }{2}x\right) \\ let{f}^{\text{'}}\left(x\right)=0 \\ \therefore \frac{\pi }{2}\cos \left(\frac{\pi }{2}x\right)=0 \\ \Rightarrow \cos \left(\frac{\pi }{2}x\right)=0 \\ \Rightarrow \frac{\pi }{2}x=(2k+1)\frac{\pi }{2}wherekisanyinteger \\ \Rightarrow x=2k+1 \\ x=2k+1 \end{gathered}$$

Step 3. Finding increasing and decreasing intervals: 

Intervals of the given function :

$$\begin{gathered} f^{\text{'}}\left(x\right)hasx=2k+2 \\ {f}^{\text{'}}(2k+2)=\frac{\pi }{2}\cos \left(\frac{\pi }{2}\right(2k+2\left)\right) \\ =\frac{\pi }{2}\cos (k+1)\pi \\ =\left\{\begin{array}{ll}-\frac{\pi }{2}& whenk=1,3,5,...\\ \frac{\pi }{2}& whenk=2,4,6,...\end{array}\right. \end{gathered}$$

f(x)is increasing on the interval $\left[2k+1,2k+3\right]$

and decreasing elsewhere.

Step 4. Verifying algebraic answers with graphs :