Question

f(x)=6x3−54x2−126x−8 is decreasing on the interval ( equation editorEquation Editor , equation editorEquation Editor ). It is increasing on the interval ( −[infinity], equation editorEquation Editor ) and the interval ( equation editorEquation Editor , [infinity] ).

Answer 1

Answer:

The function $f(x)=6x^3-54x^2-126x-8$ is decreasing on the interval $(-1,7)$ and it is increasing on the interval $(-\infty, -1)\cup (7, \infty)$

Step-by-step explanation:

To determine the intervals of increase and decrease of the function $f(x)=6x^3-54x^2-126x-8$, perform the following steps:

1. Differentiate the function

$\frac{d}{dx}\left(6x^3-54x^2-126x-8\right)=\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dx}\left(6x^3\right)-\frac{d}{dx}\left(54x^2\right)-\frac{d}{dx}\left(126x\right)-\frac{d}{dx}\left(8\right)\\\\f'(x)=18x^2-108x-126$

2. Obtain the roots of the derivative, f'(x) = 0

$\mathrm{Factor\:out\:common\:term\:}18:\quad 18\left(x^2-6x-7\right)\\\\\mathrm{Factor}\:x^2-6x-7:\quad \left(x+1\right)\left(x-7\right)\\\\18x^2-108x-126:\quad 18\left(x+1\right)\left(x-7\right)\\\\18x^2-108x-126=0\quad :\quad x=-1,\:x=7$

3.  Form open intervals with the roots of the derivative and take a value from every interval and find the sign they have in the derivative.

If f'(x) > 0, f(x) is increasing.

If f'(x) < 0, f(x) is decreasing.

On the interval $\left(-\infty, -1\right)$, take x = -2,

$f'(-2)=18(-2)^2-108(-2)-126=162$  f'(x) > 0 therefore f(x) is increasing

On the interval $\left(-1,7)$, take x = 0,

$f'(0)=18(0)^2-108(0)-126=-126$  f'(x) < 0 therefore f(x) is decreasing

On the interval $\left(7, \infty\right)$, take x = 10,

$f'(10)=18(10)^2-108(10)-126=594$  f'(x) > 0 therefore f(x) is increasing

The function $f(x)=6x^3-54x^2-126x-8$ is decreasing on the interval $(-1,7)$ and it is increasing on the interval $(-\infty, -1)\cup (7, \infty)$