Step 1
Critical points on the graph of a function where the derivative is zero or the derivative does not exist are important to consider in many application problems of the derivative. The point ( x, f(x)) is called a critical point of f(x) if x is in the domain of the function and either f′(x) = 0 or f′(x) does not exist.
Step 2
Find f'(x)
[tex]\begin{gathered} f(x)=\frac{x}{4}+\sin (\frac{x}{2}) \\ f\text{'(x)=}\frac{1}{4}(\frac{d}{dx}(x))+\frac{d}{dx}\mleft(\sin \mleft(\frac{x}{2}\mright)\mright) \end{gathered}[/tex][tex]\begin{gathered} f^{\prime}(x)=\frac{1}{4}\text{ }(1)\text{ + }\cos \mleft(\frac{x}{2}\mright)\frac{d}{dx}\mleft(\frac{x}{2}\mright) \\ f\text{'(x)=}\frac{1}{4}+\cos \mleft(\frac{x}{2}\mright)\frac{1}{2} \\ \frac{1}{4}+\cos (\frac{x}{2})\frac{1}{2}=0 \end{gathered}[/tex]
But for the critical point f'(x)=0
[tex]\begin{gathered} \frac{1}{4}+\cos (\frac{x}{2})\frac{1}{2}=0 \\ \cos (\frac{x}{2})\frac{1}{2}=-\frac{1}{4} \\ \cos (\frac{x}{2})=-\frac{1}{2} \\ Domain\text{ of }\frac{x}{4}+\sin (\frac{x}{2})\text{ is restricted to \lbrack{}0,2}\pi\rbrack \\ x=\frac{4\pi}{3} \end{gathered}[/tex]
Hence;
[tex]\begin{gathered} f(\frac{4\pi}{3})=\frac{\frac{4\pi}{3}}{4}+\sin (\frac{\frac{4\pi}{3}}{2}) \\ f(\frac{4\pi}{3})=\frac{1}{3}\pi+\frac{\sqrt[]{3}}{2}=1.913222955 \\ f(\frac{4\pi}{3})\approx1.9132\text{ to 4 decimal places} \end{gathered}[/tex]
Hence, the critical points will be;
[tex]\begin{gathered} ((\frac{4\pi}{3}),1.9132) \\ \end{gathered}[/tex]
