find any relative extrema of the function. list each extremum along with the x - value at which it occurs…

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)= - 20 - 8x - x²\nthe 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.)\nd. there are no relative minimum points and there are no relative maximum points.\nidentify 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.\na. 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.)\nb. 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.)\nc. 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.)
Answer
Explanation:
Step1: Find the derivative
The function is $g(x)=-20 - 8x - x^{2}$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, the derivative $g'(x)=\frac{d}{dx}(-20)-\frac{d}{dx}(8x)-\frac{d}{dx}(x^{2})=-8 - 2x$.
Step2: Find the critical points
Set $g'(x) = 0$. So, $-8 - 2x=0$. Solving for $x$ gives $-2x=8$, then $x=-4$.
Step3: Use the first - derivative test
Choose a test point in the interval $x\lt - 4$, say $x=-5$. Then $g'(-5)=-8-2\times(-5)=-8 + 10 = 2>0$, so the function is increasing on the interval $(-\infty,-4)$. Choose a test point in the interval $x\gt - 4$, say $x=-3$. Then $g'(-3)=-8-2\times(-3)=-8 + 6=-2<0$, so the function is decreasing on the interval $(-4,\infty)$. Since the function changes from increasing to decreasing at $x = - 4$, there is a relative maximum at $x=-4$. Find the $y$ - value by substituting $x=-4$ into $g(x)$: $g(-4)=-20-8\times(-4)-(-4)^{2}=-20 + 32-16=-4$.
Answer:
The relative maximum point is $(-4,-4)$. The function $g(x)$ is increasing over the interval $(-\infty,-4)$ and decreasing over the interval $(-4,\infty)$. So the answer for the first part is the relative maximum point $(-4,-4)$ and for the second part is B. The function $g(x)$ is increasing over the interval $(-\infty,-4)$ and decreasing over the interval $(-4,\infty)$.