How do you find the exact relative maximum and minimum of the polynomial function of #f(x)=x^3+6x^2-36x#?

1 Answer
Nallasivam V
Mar 25, 2016

The function has a relative minimum at #x=2#
The function has a relative maximum at #x=-6#

Explanation:

Given -

#y=x^3+6x^2-36x#

#dy/dx=3x^2+12x-36#
#(d^2y)/(dx^2)=6x+12#

#dy/dx=0 => 3x^2+12x-36=0#

#3x^2+12x-36=0# [Dividing both sides by 3 we get]
#x^2+4x-12=0#
#x^2+6x-2x-12=0#
#x(x+6)-2(x+6)=0#
#(x-2)(x+6)=0#

#x # has two values

#x-2=0#
#x=2#

#x+6=0#
#x=-6#

At #x=2#

#(d^2y)/(dx^2)=6(2)+12=12+12=24>0#

At #x=2#; #dy/dx=0#;#(d^2y)/(dx^2)>0#

Hence the function has a relative minimum at #x=2#

At #x=-6#

#(d^2y)/(dx^2)=6(-6)+12=-36+12=-24<0#

At #x=-6#; #dy/dx=0#;#(d^2y)/(dx^2)<0#

Hence the function has a relative maximum at #x=-6#

enter link description here

Related questions
See all questions in First Derivative Test vs Second Derivative Test for Local Extrema
Impact of this question
1648 views around the world
You can reuse this answer
Creative Commons License