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.
f(x)=-2 + 2x^3
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.)
d. 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 f(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 f(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 f(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

Differentiate $F(x)=- 2 + 2x^{3}$ using the power - rule. The derivative $F^\prime(x)=\frac{d}{dx}(-2)+\frac{d}{dx}(2x^{3})$. Since $\frac{d}{dx}(c)=0$ for a constant $c$ and $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $F^\prime(x)=6x^{2}$.

Step2: Find critical points

Set $F^\prime(x) = 0$. So, $6x^{2}=0$. Solving for $x$, we get $x = 0$.

Step3: Use the first - derivative test

Choose test points in the intervals $(-\infty,0)$ and $(0,\infty)$. Let's choose $x=-1$ for the interval $(-\infty,0)$ and $x = 1$ for the interval $(0,\infty)$.
For $x=-1$, $F^\prime(-1)=6(-1)^{2}=6>0$.
For $x = 1$, $F^\prime(1)=6(1)^{2}=6>0$.

Since the derivative $F^\prime(x)=6x^{2}\geq0$ for all real $x$ (and $F^\prime(x)$ only equals $0$ at $x = 0$), the function is increasing on $(-\infty,\infty)$.

Answer:

The relative minimum point(s) is/are none and the relative maximum point(s) is/are none.
The function $F(x)$ is increasing over the interval $(-\infty,\infty)$ and is not decreasing anywhere.