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

Explanation:

Step1: Find the first - derivative

Given $f(x)=\frac{1}{4}x^{2}-\frac{1}{2}x + \frac{13}{4}$, then $f'(x)=\frac{1}{4}\times2x-\frac{1}{2}=\frac{1}{2}x-\frac{1}{2}$.

Step2: Set the first - derivative equal to zero

Set $f'(x) = 0$, so $\frac{1}{2}x-\frac{1}{2}=0$. Add $\frac{1}{2}$ to both sides: $\frac{1}{2}x=\frac{1}{2}$, then $x = 1$.

Step3: Find the y - value

Substitute $x = 1$ into $f(x)$: $f(1)=\frac{1}{4}(1)^{2}-\frac{1}{2}(1)+\frac{13}{4}=\frac{1 - 2+13}{4}=3$. So the extreme point is $(1,3)$.

Step4: Find the second - derivative

$f''(x)=\frac{1}{2}>0$.

Step5: Determine the type of extreme point

Since $f''(1)=\frac{1}{2}>0$, the function has a relative minimum at the point $(1,3)$.

Answer:

The relative extreme point is $(1,3)$. It is a relative minimum.