Question

the graph of the following function has one relative extreme point. find it and determine whether it is a relative maximum or a relative minimum.
f(x) = 1/4 x^2 + 1/2 x + 13/4
the relative extreme point is
(type an ordered pair.)

Explanation:

Step1: Find the derivative

Given $f(x)=\frac{1}{4}x^{2}+\frac{1}{2}x+\frac{13}{4}$, using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=\frac{1}{4}\times2x+\frac{1}{2}=\frac{1}{2}x+\frac{1}{2}$.

Step2: Set the derivative equal to zero

Set $f'(x) = 0$, so $\frac{1}{2}x+\frac{1}{2}=0$. Subtract $\frac{1}{2}$ from both sides: $\frac{1}{2}x=-\frac{1}{2}$. Then multiply both sides by 2 to get $x=-1$.

Step3: Find the second - derivative

Differentiate $f'(x)=\frac{1}{2}x+\frac{1}{2}$ with respect to $x$. Since the derivative of $\frac{1}{2}x$ is $\frac{1}{2}$ and the derivative of $\frac{1}{2}$ is 0, $f''(x)=\frac{1}{2}>0$.

Step4: Find the $y$ - value

Substitute $x = - 1$ into the original function $f(x)=\frac{1}{4}x^{2}+\frac{1}{2}x+\frac{13}{4}$. Then $f(-1)=\frac{1}{4}\times(-1)^{2}+\frac{1}{2}\times(-1)+\frac{13}{4}=\frac{1}{4}-\frac{1}{2}+\frac{13}{4}=\frac{1 - 2+13}{4}=3$.

Answer:

$(-1,3)$