How do you find the local max and min for # 2x^3-3x^2-36x-3#?

Here's the process, numbers and all:

Find #f'(x)# using the power rule.

#f(x)=2x^3-3x^2-36x-3#

#f'(x)=6x^2-6x-36#

The local minima and maxima will occur when the derivative equals #0#:

#6x^2-6x-36=0#

#x^2-x-6=0#

#(x-3)(x+2)=0#

#x=3" ",""" "x=-2#

These are the two points at which maxima/minima could occur. There are three ways we can identify which is which:

The First Derivative Test:

Evaluate the sign of the first derivative around each local extremum.

Around #x=-2#:

#f'(-3)=36larr"increasing"#
#f'(-1)=-24larr"decreasing"#

Since the first derivative switches from increasing to decreasing, there is a local maximum at #x=-2#.

Around #x=3#:

#f'(2)=-24larr"decreasing"#
#f'(4)=36larr"increasing"#

Since the first derivative switches from decreasing to increasing, there is a local minimum at #x=3#.

The Second Derivative Test:

Find the second derivative of the function. Then, find the sign of it at each point.

• If #f''(a)<0# and #f'(a)=0#, then there is a local maximum at #x=a#.

• If #f''(a)>0# and #f'(a)=0#, then there is a local minimum at #x=a#.

#f'(x)=6x^2-6x-36#

#f''(x)=12x-6#

Determine the sign at each extremum.

#f''(-2)=-30#

Since this is #<0#, there is a local maximum at #x=-2#.

#f''(3)=30#

Since this is #>0#, there is a local minimum at #x=3#.

Check a graph of the original function:

This method shouldn't be used as proof on a test, say, but it's a fine way to make sure you're on the right track with your answer.

graph{2x^3-3x^2-36x-3 [-5, 7, -120, 150]}

Indeed, there is a local maximum at #x=-2# and a local minimum at #x=3#. Note that these are the locations of the extrema. The actual extrema are the function values of the points, that is, #f(-2)# and #f(3)#.

Thus, the local maximum is #f(-2)=41# and the local minimum is #f(3)=-84#.