the graph of the following function has one relative extreme point. find it and determine whether it is a…

Explanation:

Step1: Find the derivative

Given $f(x)=\frac{1}{4}x^{2}-4$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=\frac{1}{4}\times2x=\frac{1}{2}x$.

Step2: Find the critical points

Set $f'(x) = 0$. So, $\frac{1}{2}x=0$, which gives $x = 0$.

Step3: Find the y - coordinate of the critical point

Substitute $x = 0$ into the original function $f(x)$. Then $f(0)=\frac{1}{4}(0)^{2}-4=-4$. The relative extreme point is $(0,-4)$.

Step4: Determine if it's a max or min

Find the second - derivative $f''(x)$. Since $f'(x)=\frac{1}{2}x$, then $f''(x)=\frac{1}{2}>0$. When $f''(x)>0$ at a critical point, the function has a relative minimum at that point.

Answer:

The relative extreme point is $(0, - 4)$. It is a relative minimum.