How do you find the exact relative maximum and minimum of the polynomial function of #f(x)=2x^3+3x^2-12x #?

1 Answer
Feb 5, 2016

Relative maximum: #f(-2)=20#
Relative minimum: #f(1) = -9#

Explanation:

Given
#color(white)("XXX")f(x)=2x^3+3x^2-12x#

Note
#color(white)("XXX")#Relative minimums/maximums happen points where #f'(x)=0#

#f'(x) = 6x^2+6x-12#
#color(white)("XXX")=6(x^2+x-2)#
#color(white)("XXX")=6(x+2)(x-1)#

#f'(x)=0# for # x=-2# and # x=+1#
so these are the locations of the relative minimum/maximum values.

#f(-2) = -16+12+24=20#
#f(1) = 2+3-12 = -9#

Therefore
#color(white)("XXX")#the relative maximum is #20# (at #x=-2#)
#color(white)("XXX")#the relative minimum is #(-9)# ( at #x=1#)