How do you find the local max and min for 2x^3-3x^2-36x-32x33x236x3?

1 Answer
Jan 31, 2016

Local maximum of 4141 at x=-2x=2 and local minimum of -8484 at x=3x=3.

Explanation:

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.