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

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.\nf(x) = 1/4 x^2 + 1/2 x + 13/4\nthe relative extreme point is \n(type an ordered pair.)

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.\nf(x) = 1/4 x^2 + 1/2 x + 13/4\nthe relative extreme point is \n(type an ordered pair.)

Answer

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:

$(-1,3)$