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#

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