How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for #f(x)=x^3-3x^2+1#?

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Answer 1 · 2016-12-08T00:47:49

At #x=0# The function has a relative maximum.

At #x=2# the function has a relative minimum.

Given -

#y=x^3-3x^2+1#

#dy/dx=3x^2-6x#

#(d^2y)/(dx^2)=6x-6#

At any given point, if #dy/dx>0# the function is increasing otherwise it is decreasing.

#dy/dx=0 => 3x^2-6x#
#3x(x-2)=0#

#3x=0#
#x=0#

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

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

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

The function has a relative maximum.

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

At #x=2; dy/dy =0;(d^2y)/(dx^2)>0#
There is a relative minimum.

graph{x^3-3x^2+1 [-10, 10, -5, 5]}