Question

use a graphing calculator to graph the function g(x)=x^3 + 4x^2 + 1. over what interval is the function increasing? (select all that apply.) -∞ < x < -2.7 -2.7 < x < 1.8 -∞ < x < 0 1.8 < x < ∞ 0 < x < ∞ 4 < x < 0

Explanation:

Step1: Find the derivative of the function.

The derivative of $g(x)=x^{3}+4x^{2}+1$ is $g'(x)=3x^{2}+8x$ using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$.

Step2: Set the derivative equal to zero and solve for $x$.

$3x^{2}+8x = 0$. Factor out an $x$: $x(3x + 8)=0$. So $x = 0$ or $x=-\frac{8}{3}\approx - 2.7$.

Step3: Test intervals.

We have three intervals to test: $(-\infty,-\frac{8}{3})$, $(-\frac{8}{3},0)$ and $(0,\infty)$.

• For $x<- \frac{8}{3}$, let $x=-3$. Then $g'(-3)=3\times(-3)^{2}+8\times(-3)=27 - 24 = 3>0$.
• For $-\frac{8}{3}
• For $x>0$, let $x = 1$. Then $g'(1)=3\times1^{2}+8\times1=3 + 8 = 11>0$.

The function $g(x)$ is increasing when $g'(x)>0$. So the function is increasing on the intervals $-\infty

Answer:

$-\infty