Max and min of f(x) = 2x^3 + 3x^2 - 12x using the second derivative

Step by step

The question

Question: find the maximum and minimum points of f(x) = 2x³ + 3x² − 12x with the second derivative test.

1 Find f′(x)

Use the power rule on each term.

Now we have f′(x) = 6x² + 6x − 12.

2 Solve f′(x) = 0

At a turning point the graph is flat for a moment: the slope is 0.

So we solve f′(x) = 0.

Take out the 6.

We need two numbers that multiply to −2 and add to 1: 2 and −1.

A product is 0 only when one of its parts is 0.

So x = −2 or x = 1.

3 Find f″(x)

f double prime is the derivative of f′.

It tells how the slope changes.

Use the power rule again, now on f′.

Now we have f″(x) = 12x + 6.

If f″ is positive, the slope is growing: the curve bends up, like a cup.

Its bottom is a minimum.

If f″ is negative, it bends down, like a cap.

Its top is a maximum.

4 Test each point

At x = −2, f″ is −18: negative.

A cap.

So a maximum.

Careful!

Negative f″ is a cap.

The top of a cap is a maximum, not a minimum.

At x = 1, f″ is 18: positive.

A cup.

So a minimum.

Its y: put −2 into f.

We get 20.

Its y: put 1 into f.

We get −7.

So: a maximum at (−2, 20), and a minimum at (1, −7).

5 Check

Let's check with the graph.

A cap at the maximum, and a cup at the minimum.

Right!

Let's look back.

First: find f′(x).

Then: solve f′(x) = 0.

Then: find f″(x).

Then: test each point.

Last: check.