# How do you find the critical numbers of f(x)=(x+1)^7-7x-2

*print*Print*list*Cite

### 1 Answer

`f(x)=(x+1)^7-7x-2`

To determine the critical numbers, we have to take the derivative of the function.

`f'(x)=7(x +1)^6 - 7 -0`

`f'(x) = 7(x+1)^6-7`

Then, set the derivative equal to zero.

0=7(x+1)^6-7

To solve for the value of x, factor the right side.

`0=7[(x+1)^6-1]`

`0=7{[(x+1)^3-1][(x+1)^3+1]}`

Simplify the equation by dividing both sides by 7.

`0=[(x+1)^3-1][(x+1)^3+1]`

Then, set each factor equal to zero. And isolate the x.

For the first factor:

`(x+1)^3-1=0`

`(x+1)^3=1`

`x+1=root(3)(1)`

`x+1=1`

`x=0`

For the second factor:

`(x+1)^3+1=0`

`(x+1)^3=-1`

`x+1=root(3)(-1)`

`x+1=-1`

`x=-2`

**Therefore, the critical numbers are `x=-2` and `x=0` .**