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)=\frac{1}{4}x^{2}+x + 2\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)=\frac{1}{4}x^{2}+x + 2\nthe relative extreme point is \n(type an ordered pair.)

Answer

Explanation:

Step1: Find the derivative

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

Step2: Find the critical points

Set (f^\prime(x) = 0), so (\frac{1}{2}x+1 = 0). Solving for (x): (\frac{1}{2}x=-1), then (x=-2).

Step3: Find the second - derivative

Differentiate (f^\prime(x)) with respect to (x). Since (f^\prime(x)=\frac{1}{2}x + 1), then (f^{\prime\prime}(x)=\frac{1}{2}>0).

Step4: Find the (y) - coordinate of the extreme point

Substitute (x = - 2) into (f(x)): (f(-2)=\frac{1}{4}\times(-2)^{2}+(-2)+2=\frac{1}{4}\times4=1).

Answer:

((-2,1)) (a relative minimum since (f^{\prime\prime}(x)>0) at (x=-2))