Question

find any relative extrema of the function. list each extremum along with the x - value at which it occurs. identify intervals over which the function is increasing and over which it is decreasing. then sketch a graph of the function.
g(x)=-22 + 12x-2x^{2}
the relative minimum point(s) is/are and there are no relative maximum points
(simplify your answer. type an ordered pair, using integers or fractions. use a comma to separate answers as needed.)
there are no relative minimum points and there are no relative maximum points
identify any intervals over which the function is increasing or decreasing. select the correct choice below and fill in the answer box(es) within your choice.
a. the function g(x) is increasing over the interval(s) and decreasing over the interval(s)
(type your answers in interval notation. use integers or fractions for any numbers in the expressions. use a comma to separate answers as needed.)
b. the function g(x) is increasing over the interval(s) and is not decreasing anywhere.
(type your answer in interval notation. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)
c. the function g(x) is decreasing over the interval(s) and is not increasing anywhere.
(type your answer in interval notation. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)

Explanation:

Step1: Find the derivative of the function

The derivative of $g(x)=-22 + 12x-2x^{2}$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $g'(x)=12-4x$.

Step2: Find the critical points

Set $g'(x) = 0$. So, $12-4x=0$. Solving for $x$ gives $4x = 12$, then $x = 3$.

Step3: Determine the nature of the critical point

Take the second - derivative of $g(x)$. $g''(x)=\frac{d}{dx}(12 - 4x)=-4$. Since $g''(3)=-4<0$, the function has a relative maximum at $x = 3$.
Substitute $x = 3$ into $g(x)$: $g(3)=-22+12\times3-2\times3^{2}=-22 + 36-18=-4$. So the relative maximum is $(-4)$ at $x = 3$.

Step4: Find the intervals of increase and decrease

Set $g'(x)>0$ to find where the function is increasing:
$12-4x>0$, $4x<12$, $x < 3$. So the function is increasing on the interval $(-\infty,3)$.
Set $g'(x)<0$ to find where the function is decreasing:
$12-4x<0$, $4x>12$, $x>3$. So the function is decreasing on the interval $(3,\infty)$.

Answer:

The relative maximum point is $(3,-4)$.
A. The function $g(x)$ is increasing over the interval $(-\infty,3)$ and decreasing over the interval $(3,\infty)$.