Turning points of f(x) = x^3 - 3x^2

Step by step

The question

Question: find the maximum and minimum points of f(x) = x³ − 3x² − 9x.

f(x) = x³ − 3x² − 9x

The derivative is the slope.

Where it is positive, the graph goes up.

Where it is negative, the graph goes down.

f′(x) > 0: going up ↗

f′(x) < 0: going down ↘

1 Find f′(x)

Use the power rule on each term.

f′(x) = 3x^(3 − 1) − 3 × 2x^(2 − 1) − 9 × 1

Now we have f′(x) = 3x² − 6x − 9.

= 3x² − 6x − 9

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.

f′(x) = 0

3x² − 6x − 9 = 0

Take out the 3.

3(x² − 2x − 3) = 0

We need two numbers that multiply to −3 and add to −2: 1 and −3.

3(x + 1)(x − 3) = 0

1 × (−3) = −3, 1 + (−3) = −2

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

a × b = 0 → a = 0 or b = 0

So x = −1 or x = 3.

x = −1 or x = 3

Careful!

f′ = 0 alone does not tell max or min.

The signs on each side tell us.

Careful: f′(x) = 0 is not enough: look at the signs

3 Sign table

A sign table.

The numbers −1 and 3 cut the line into 3 parts.

Test one number in each part: −2, 0, 4.

f′(−2) = 15, f′(0) = −9, f′(4) = 15

f′(−2) is 15: positive.

So f goes up there.

f′(0) is −9: negative.

So f goes down there.

f′(4) is 15: positive.

So f goes up there.

4 Max and min

At x = −1, f goes up, then down.

That is a maximum.

max

At x = 3, f goes down, then up.

That is a minimum.

min

Its y: put −1 into f.

We get 5.

f(−1) = (−1)³ − 3(−1)² − 9(−1) = −1 − 3 + 9 = 5

Its y: put 3 into f.

We get −27.

f(3) = 3³ − 3(3)² − 9(3) = 27 − 27 − 27 = −27

So: a maximum at (−1, 5), and a minimum at (3, −27).

= max (−1, 5), min (3, −27)

5 Check

Let's check with the graph.

Up, then down, then up: just like the table.

Right!

Let's look back.

First: find f′(x).

Then: solve f′(x) = 0.

Then: sign table.

Then: max and min.

Last: check.

Answer