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

1 Answer
Nov 14, 2016

Answer:

So the critical points are #(-1,20)# and #(1,-7)#

#(-1,20)# is a maximum
#(1,-7)# is a minium

Explanation:

We have # f(x) = 2x^3 + 3x^2 - 12x #

To identify the critical vales, we differentiate and find find values of #x# st #f'(x)=0#

# { ( f'(x) < 0, => f(x) " is decreasing" ), ( f'(x) = 0, => f(x) " is stationary" ), ( f'(x) > 0, => f(x) " is increasing" ) :} #

Differentiating wrt #x#' we have:

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

At a critical point, # f'(x)=0 #

# 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 #f(x)#
# x=-2 => f(-2)=2(-8)+3(4)-12(-2) = -16+12+24=20 #
# x=1 => f(1)=2+3-12=-7 #

So the critical points are #(-1,20)# and #(1,-7)#

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

# { ( f''(x) < 0, => f'(x) " is decreasing" => "maximum" ), ( f''(x) = 0, => f'(x) " is stationary" => "inflection" ), ( f''(x) > 0, => f'(x) " is increasing" => "minimum" ) :} #

Differentiating [1] wrt #x# gives s;

# f''(x) = 12x + 6 #
# x=-2 => f''(-2)=-14+6 < 0 #, ie a maximum
# x=1 => f''(1)= 12+6>0#, ie a minimum

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Incidental, As this is a cubic with a positive coefficient of #x^3#, we can deduce that the critical point corresponding to the smallest value of #x# must be the maximum and that corresponding to the larger value must be the maximum. It is not possible to have any other possibility!