How do you locate the critical points of the function #f(x) = x^3 - 15x^2 + 4# and use the Second Derivative Test to determine whether they correspond to local maxima, local minima, or neither?

1 Answer
Mar 26, 2015

Critical point: number #c# in the domain of #f# with #f'(c)=0# or #f'(c) does not exist.

Finding Critical Points
#f(x)=x^3-15x^2+4#

#f'(x)=3x^2-30x#

#f'(x)# does not exist -- no such #x#

#f'(x)=3x^2 - 30x=0#
#3x(x-10)=0# at #x=0#, #x=10#

Both #0# and #10# are in the domain of #f#, so they are both crical points for #f#

Testing Critical Points
The second derivative of #f#:

#f''(x)=6x-30#

At the critical point #0#, we have #f''(0)=-30 <0#
The second derivative test (for local extrema) tells us that, #f(0)# is a local maximum. There is a local maximum of #4# at #0#..

At the critical point #10#, we have #f''(10)=6(10)-30 > 0#
The second derivative test (for local extrema) tells us that, #f(10)# is a local minimum. There is a local minimum of #-496# at #10#..

And here's the graph (you'll have to zoom to see details):

graph{y=x^3-15x^2+4 [-16, 41.74, -19.96, 8.92]} #