Question

for the following function, a) give the coordinates of any critical points and classify each point as a relative maximum, a relative minimum, or neither, b) identify intervals where the function is increasing or decreasing, c) give the coordinates of any points of inflection, d) identify intervals where the function is concave up or concave down, and e) sketch the graph. h(x)=3x^3 - 9x (simplify your answer. use integers or fractions for any numbers in the expression. type an ordered - pair. use a comma to separate answers as needed.) d there are no relative minimum points and there are no relative maximum points b) on what interval(s) is h increasing or decreasing? select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. a the function is increasing on (-∞, - 1), (1,∞). the function is decreasing on (-1,1) (simplify your answers. type your answers in interval notation. use a comma to separate answers as needed.) b the function is decreasing on the function is never increasing (simplify your answer. type your answer in interval notation. use a comma to separate answers as needed.) c the function is increasing on the function is never decreasing (simplify your answer. type your answer in interval notation. use a comma to separate answers as needed.) d the function is never increasing or decreasing c) what are the coordinates of the inflection point(s)? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a the coordinates of the inflection point(s) are (simplify your answer. type an ordered pair. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.) b there are no inflection points

Explanation:

Step1: Find the first - derivative

Given $h(x)=3x^{3}-9x$, using the power rule $(x^n)^\prime = nx^{n - 1}$, we have $h^\prime(x)=9x^{2}-9$.

Step2: Find critical points

Set $h^\prime(x) = 0$, so $9x^{2}-9 = 0$. Factor out 9: $9(x^{2}-1)=0$, then $x^{2}-1=(x - 1)(x + 1)=0$. Solving gives $x=-1,1$.
Substitute $x=-1$ into $h(x)$: $h(-1)=3(-1)^{3}-9(-1)=-3 + 9 = 6$.
Substitute $x = 1$ into $h(x)$: $h(1)=3(1)^{3}-9(1)=3-9=-6$.
The critical points are $(-1,6)$ and $(1,-6)$.
To classify, we use the second - derivative test.

Step3: Find the second - derivative

$h^{\prime\prime}(x)=18x$.
For $x=-1$, $h^{\prime\prime}(-1)=18(-1)=-18<0$, so $(-1,6)$ is a relative maximum.
For $x = 1$, $h^{\prime\prime}(1)=18(1)=18>0$, so $(1,-6)$ is a relative minimum.

Step4: Find intervals of increase and decrease

$h^\prime(x)=9x^{2}-9=9(x^{2}-1)$.
$h^\prime(x)>0$ when $x^{2}-1>0$, i.e., $x<-1$ or $x>1$. So the function is increasing on $(-\infty,-1)\cup(1,\infty)$.
$h^\prime(x)<0$ when $x^{2}-1<0$, i.e., $-1

Step5: Find inflection points

Set $h^{\prime\prime}(x)=0$, so $18x = 0$, which gives $x = 0$.
Substitute $x = 0$ into $h(x)$: $h(0)=3(0)^{3}-9(0)=0$.
The inflection point is $(0,0)$.

Step6: Find concavity

$h^{\prime\prime}(x)=18x$.
$h^{\prime\prime}(x)>0$ when $x>0$, so the function is concave up on $(0,\infty)$.
$h^{\prime\prime}(x)<0$ when $x<0$, so the function is concave down on $(-\infty,0)$.

Answer:

a) Relative maximum point: $(-1,6)$; Relative minimum point: $(1,-6)$
b) The function is increasing on $(-\infty,-1),(1,\infty)$; The function is decreasing on $(-1,1)$
c) The coordinates of the inflection point are $(0,0)$
d) The function is concave up on $(0,\infty)$; The function is concave down on $(-\infty,0)$
e) To sketch the graph: Mark the relative maximum at $(-1,6)$, relative minimum at $(1,-6)$, and inflection point at $(0,0)$. The function is increasing on $(-\infty,-1)$ and $(1,\infty)$, decreasing on $(-1,1)$, concave down on $(-\infty,0)$ and concave up on $(0,\infty)$.