How do you locate the critical points of the function f(x) = x^3 - 15x^2 + 4f(x)=x3−15x2+4 and use the Second Derivative Test to determine whether they correspond to local maxima, local minima, or neither?

1 Answer
Jim H
Mar 26, 2015

Critical point: number cc in the domain of ff 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]} #

Related questions
See all questions in Critical Points of Inflection
Impact of this question
5520 views around the world
You can reuse this answer
Creative Commons License