use a graphing calculator to graph the function g(x)=x^3 + 4x^2 + 1. over what interval is the function…

Explanation:

Step1: Find the derivative

The derivative of $g(x)=x^{3}+4x^{2}+1$ is $g'(x)=3x^{2}+8x$.

Step2: Set derivative equal to 0

Set $g'(x) = 0$, so $3x^{2}+8x=0$. Factor out $x$: $x(3x + 8)=0$. The solutions are $x = 0$ and $x=-\frac{8}{3}\approx - 2.7$.

Step3: Test intervals

We have three intervals to test: $(-\infty,-\frac{8}{3})$, $(-\frac{8}{3},0)$ and $(0,\infty)$. For $x\in(-\infty,-\frac{8}{3})$, let $x=-3$. Then $g'(-3)=3\times(-3)^{2}+8\times(-3)=27 - 24 = 3>0$. For $x\in(-\frac{8}{3},0)$, let $x=-1$. Then $g'(-1)=3\times(-1)^{2}+8\times(-1)=3 - 8=-5<0$. For $x\in(0,\infty)$, let $x = 1$. Then $g'(1)=3\times1^{2}+8\times1=3 + 8 = 11>0$. The function is increasing on $(-\infty,-\frac{8}{3})\cup(0,\infty)$. Among the given options, the correct ones are $-\infty<x<0$ and $-2.7<x<1.8$ (since $-2.7\approx-\frac{8}{3}$ and the interval $-2.7<x<1.8$ overlaps with the increasing - intervals).

Answer:

-2.7 < x < 1.8, -∞ < x < 0