# How do you find the critical numbers of #f(x)= 2x^3 + 3x^2-12x#?

##### 1 Answer

So the critical points are

#### Explanation:

We have

To identify the critical vales, we differentiate and find find values of

Differentiating wrt

# f'(x) = 6x^2 + 6x - 12 # .... [1]

At a critical point,

# f'(x)=0 => 6x^2 + 6x - 12 = 0 #

# :. x^2 + x - 2 = 0 #

# :. (x+2)(x-1) = 0 #

# x=-2,1 #

Ton find the y-coordinate we substitute the required value into

So the critical points are

Although this answers the question, let's go a bit further and identify the nature of these critical points by looking at the sign of second derivative, and

Differentiating [1] wrt

# f''(x) = 12x + 6 #

# x=-2 => f''(-2)=-14+6 < 0 # , ie a maximum

# x=1 => f''(1)= 12+6>0# , ie a minimum

Incidental, As this is a **cubic** with a **positive coefficient** of