`f(x) = -2x^3 + 3x^2 - 12x` Find the critical numbers, open intervals on which the function is increasing or decreasing, apply first derivative test to identify all relative extrema.

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Chapter 3, 3.3 - Problem 21 - Calculus of a Single Variable (10th Edition, Ron Larson).
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tiburtius | High School Teacher | (Level 2) Educator

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Critical points are where derivation is equal to zero i.e.

`f'(x)=0`

Therefore, we will calculate function derivative.

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

Hence we need to find solution to the following equation.

`-6x^2+6x-12=0`

Divide by -6.

`x^2-x+2=0`

Apply quadratic formula `x_(1,2)=(-b pm sqrt(b^2-4ac))/(2a)`

`x_(1,2)=(1 pm sqrt(1-8))/2=(1 pm i sqrt(7))/2`

The above equation has no real roots meaning that the function has no critical points. Hence the function is increasing or decreasing over the whole domain. To check whether it is increasing or decreasing we only need to check sign of the derivative at any point (any point will do because the function is monotone over whole the domain).

`f'(0)=-12`

Since the derivative is negative the function is decreasing over whole the domain i.e. `(-oo,+oo).`

Graph of the function is shown on the image below.

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