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

Step by step

The question

Question: find the maximum and minimum points of f(x) = 2x³ − 6x.

f(x) = 2x³ − 6x

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) = 2 × 3x^(3 − 1) − 6 × 1

Now we have f′(x) = 6x² − 6.

= 6x² − 6

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

6x² − 6 = 0

Take out the 6.

6(x² − 1) = 0

x² − 1 is a difference of two squares: x² minus 1 squared.

So it is (x − 1)(x + 1).

6(x + 1)(x − 1) = 0

a² − b² = (a − b)(a + b), b = 1

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 = 1.

x = −1 or x = 1

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 1 cut the line into 3 parts.

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

f′(−2) = 18, f′(0) = −6, f′(2) = 18

f′(−2) is 18: positive.

So f goes up there.

f′(0) is −6: negative.

So f goes down there.

f′(2) is 18: 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 = 1, f goes down, then up.

That is a minimum.

min

Its y: put −1 into f.

We get 4.

f(−1) = 2(−1)³ − 6(−1) = −2 + 6 = 4

Its y: put 1 into f.

We get −4.

f(1) = 2(1)³ − 6(1) = 2 − 6 = −4

So: a maximum at (−1, 4), and a minimum at (1, −4).

= max (−1, 4), min (1, −4)

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