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.
$f(x)=\frac{1}{4}x^{2}+2x + 6$
the relative extreme point is
(type an ordered pair.)
is the relative extreme point a relative maximum or a relative minimum?
relative minimum
relative maximum

Explanation:

Step1: Find the first - derivative

The function is \(f(x)=\frac{1}{4}x^{2}+2x + 6\).
Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we have \(f^\prime(x)=\frac{1}{4}\times2x+2=\frac{1}{2}x + 2\).

Step2: Find the critical point

Set \(f^\prime(x)=0\), so \(\frac{1}{2}x+2 = 0\).
Subtract 2 from both sides: \(\frac{1}{2}x=-2\).
Multiply both sides by 2: \(x=-4\).

Step3: Find the second - derivative

Differentiate \(f^\prime(x)=\frac{1}{2}x + 2\) with respect to \(x\).
Using the power rule, \(f^{\prime\prime}(x)=\frac{1}{2}\).

Step4: Evaluate the function at the critical point

Substitute \(x = - 4\) into \(f(x)\):
\(f(-4)=\frac{1}{4}\times(-4)^{2}+2\times(-4)+6\).
First, \(\frac{1}{4}\times(-4)^{2}=\frac{1}{4}\times16 = 4\), \(2\times(-4)=-8\).
Then \(f(-4)=4-8 + 6=2\).

Since \(f^{\prime\prime}(x)=\frac{1}{2}>0\), the function is concave up at \(x=-4\).

Answer:

The relative extreme point is \((-4,2)\).
The relative extreme point is a relative minimum.