Question

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

Answer 1

Local maximum of `41` at `x=-2` and local minimum of `-84` at `x=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`.