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

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 \(x=-2\))