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

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)`

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

`(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'\\\\(d)/(dx)\left(6x^3\right)-(d)/(dx)\left(54x^2\right)-(d)/(dx)\left(126x\right)-(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)`