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})
describe any relative extrema. select the correct choice below and if necessary, fill in the answer box(es) to within your choice.
a. the relative minimum point(s) is/are and the relative maximum point(s) is/are
(simplify your answers. type ordered - pairs, using integers or fractions. use a comma to separate answers as needed.)
b. the relative maximum point(s) is/are and there are no relative minimum points.
(simplify your answer. type an ordered - pair, using integers or fractions. use a comma to separate answers as needed.)
c. 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.)
d. there are no relative minimum points and there are no relative maximum points.

Explanation:

Step1: Find the derivative

The function is $g(x)=22 + 12x-2x^{2}$. Using the power - rule for differentiation $\frac{d}{dx}(x^{n})=nx^{n - 1}$, the derivative $g'(x)=\frac{d}{dx}(22)+\frac{d}{dx}(12x)-\frac{d}{dx}(2x^{2})=0 + 12-4x=12 - 4x$.

Step2: Find the critical points

Set $g'(x)=0$. So, $12 - 4x = 0$. Solving for $x$:
\[

$$\begin{align*} 12-4x&=0\\ -4x&=-12\\ x& = 3 \end{align*}$$

\]

Step3: Determine the second - derivative

Differentiate $g'(x)=12 - 4x$ with respect to $x$. The second - derivative $g''(x)=\frac{d}{dx}(12 - 4x)=-4$. Since $g''(3)=-4<0$, the function has a relative maximum at $x = 3$.

Step4: Find the value of the relative maximum

Substitute $x = 3$ into the original function $g(x)=22 + 12x-2x^{2}$. Then $g(3)=22+12\times3-2\times3^{2}=22 + 36-18=40$. So the relative maximum point is $(3,40)$ and there are no relative minimum points.

Step5: Determine the intervals of increase and decrease

Choose a test point in the interval $(-\infty,3)$, say $x = 0$. Then $g'(0)=12-4\times0 = 12>0$, so the function is increasing on the interval $(-\infty,3)$.
Choose a test point in the interval $(3,\infty)$, say $x = 4$. Then $g'(4)=12-4\times4=12 - 16=-4<0$, so the function is decreasing on the interval $(3,\infty)$.

Answer:

B. The relative maximum point(s) is (3,40) and there are no relative minimum points.