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

Question

Emilia Aguilar · Accepted Answer

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

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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Evelyn Agard · Answer

undefined To find the exact relative maximum and minimum of the polynomial function $ f(x) = x^3 + 6x^2 - 36x $, follow these steps:

1. Find the derivative of the function: $ f'(x) = 3x^2 + 12x - 36 $.
2. Set the derivative equal to zero and solve for $ x $ to find critical points: $ 3x^2 + 12x - 36 = 0 $.
3. Solve the quadratic equation to find the critical points: $ x = -6 $ and $ x = 2 $.
4. Evaluate the function at the critical points and at the endpoints of the interval of interest, if any.
5. The maximum and minimum values correspond to the highest and lowest function values obtained in step 4, respectively.