What are the global and local extrema of #f(x)=8x^3-4x^2+6# ?

1 Answer
Oct 30, 2016

Answer:

The local extrema are #(0,6)# and #(1/3,158/27)#
and the global extrema are #+-oo#

Explanation:

We use #(x^n)'=nx^(n-1)#
Let us find the first derivative

#f'(x)=24x^2-8x#
For local extrema #f'(x)=0#
So #24x^2-8x=8x(3x-1)=0#

#x=0# and #x=1/3#

So let's do a chart of signs
#x##color(white)(aaaaa)##-oo##color(white)(aaaaa)##0##color(white)(aaaaa)##1/3##color(white)(aaaaa)##+oo#
#f'(x)##color(white)(aaaaa)##+##color(white)(aaaaa)##-##color(white)(aaaaa)##+#
#f(x)##color(white)(aaaaaa)##uarr##color(white)(aaaaa)##darr##color(white)(aaaaa)##uarr#

So at the point #(0,6)# we have a local maximum
and at #(1/3,158/27)#
We have a point a point of inflexion #f''(x)=48x-8#
#48x-8=0##=>##x=1/6#
limit#f(x)=-oo#
#xrarr-oo#
limit#f(x)=+oo#
#xrarr+oo#
graph{8x^3-4x^2+6 [-2.804, 3.19, 4.285, 7.28]}