Question

1. describe the increasing, decreasing intervals

you may graph in desmos.
( look from left to right )
$y=-x^{3}+x^{2}+16x - 16$
$f(x)$ is decreasing
in what x interval?
$f(x)$ is constant
in what x interval?
$f(x)$ is increasing
in what x interval?

Explanation:

Step1: Find the derivative

Use the power rule \((x^n)^\prime=nx^{n - 1}\).
For \(y=-x^{3}+x^{2}+16x - 16\), the derivative \(y^\prime=f^\prime(x)=-3x^{2}+2x + 16\).

Step2: Find the critical points

Set \(y^\prime = 0\), so \(-3x^{2}+2x + 16=0\).
Multiply through by \(- 1\) to get \(3x^{2}-2x - 16=0\).
Factor: \(3x^{2}-2x - 16=(3x - 8)(x + 2)=0\).
Solve \((3x - 8)(x + 2)=0\), we get \(x=-2\) or \(x=\frac{8}{3}\).

Step3: Test intervals

• For \(x\lt - 2\), let \(x=-3\). Then \(y^\prime=-3(-3)^{2}+2(-3)+16=-27-6 + 16=-17\lt0\).
• For \(-2\lt x\lt\frac{8}{3}\), let \(x = 0\). Then \(y^\prime=-3(0)^{2}+2(0)+16=16\gt0\).
• For \(x\gt\frac{8}{3}\), let \(x = 3\). Then \(y^\prime=-3(3)^{2}+2(3)+16=-27 + 6+16=-5\lt0\).

Answer:

• \(f(x)\) is decreasing in the interval \((-\infty,-2)\cup(\frac{8}{3},\infty)\).
• \(f(x)\) is not constant in any interval.
• \(f(x)\) is increasing in the interval \((-2,\frac{8}{3})\).