Question

The question is attached in photoFunction: f(x) = x/4 + sin x/2

Answer 1

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)

`\begin{gathered} f(x)=(x)/(4)+\sin ((x)/(2)) \\ f\text{'(x)=}(1)/(4)((d)/(dx)(x))+(d)/(dx)\mleft(\sin \mleft((x)/(2)\mright)\mright) \end{gathered}`
`\begin{gathered} f^(\prime)(x)=(1)/(4)\text{ }(1)\text{ + }\cos \mleft((x)/(2)\mright)(d)/(dx)\mleft((x)/(2)\mright) \\ f\text{'(x)=}(1)/(4)+\cos \mleft((x)/(2)\mright)(1)/(2) \\ (1)/(4)+\cos ((x)/(2))(1)/(2)=0 \end{gathered}`

But for the critical point f'(x)=0

`\begin{gathered} (1)/(4)+\cos ((x)/(2))(1)/(2)=0 \\ \cos ((x)/(2))(1)/(2)=-(1)/(4) \\ \cos ((x)/(2))=-(1)/(2) \\ Domain\text{ of }(x)/(4)+\sin ((x)/(2))\text{ is restricted to \lbrack{}0,2}\pi\rbrack \\ x=(4\pi)/(3) \end{gathered}`

Hence;

`\begin{gathered} f((4\pi)/(3))=((4\pi)/(3))/(4)+\sin (((4\pi)/(3))/(2)) \\ f((4\pi)/(3))=(1)/(3)\pi+\frac{\sqrt[]{3}}{2}=1.913222955 \\ f((4\pi)/(3))\approx1.9132\text{ to 4 decimal places} \end{gathered}`

Hence, the critical points will be;

`\begin{gathered} (((4\pi)/(3)),1.9132) \\ \end{gathered}`