You need to perform first derivative test to find what are the critical points and the intervals where the function increases or decreases.

`f'(x) = ((x^2+10x+12)'*(x-2) - (x^2+10x+12)*(x-2)')/((x-2)^2)`

`f'(x) = ((2x+10)*(x-2) - (x^2+10x+12))/((x-2)^2)`

You need to open the brackets to numerator such that:

`f'(x) = (2x^2 + 6x - 20...

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You need to perform first derivative test to find what are the critical points and the intervals where the function increases or decreases.

`f'(x) = ((x^2+10x+12)'*(x-2) - (x^2+10x+12)*(x-2)')/((x-2)^2)`

`f'(x) = ((2x+10)*(x-2) - (x^2+10x+12))/((x-2)^2)`

You need to open the brackets to numerator such that:

`f'(x) = (2x^2 + 6x - 20 - x^2 - 10x - 12)/((x-2)^2)`

`f'(x) = (x^2- 4x - 32)/((x-2)^2)`

You need to solve the equation f'(x) = 0 such that:

`(x^2 - 4x - 32)/((x-2)^2) = 0 =gt x^2 - 4x - 32 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (4 +- sqrt(16 + 128))/2`

`x_(1,2) = (4 +-12)/2 =gt x_1 = 8`

`x_2 = -4`

You need to perform the sign test, hence, the derivative is positive if `x in (-oo,-4)U(8,oo)` and derivative is negative if `x in (-4,8)` .

**Hence, the function increases over `(-oo,-4)U(8,oo)` and it decreases if `x in (-4,8)` , thus, the function reaches its maximum at x=-4 and it reaches its minimum at x=8.**