Turning points of f(x) = x^3 - 12x Step by Step

Step by step

The question

Question: find the maximum and minimum points of f(x) = x³ − 12x.

f(x) = x³ − 12x

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) − 12 × 1

Now we have f′(x) = 3x² − 12.

= 3x² − 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.

f′(x) = 0

3x² − 12 = 0

Take out the 3.

3(x² − 4) = 0

x² − 4 is a difference of two squares: x² minus 2 squared.

So it is (x − 2)(x + 2).

3(x + 2)(x − 2) = 0

a² − b² = (a − b)(a + b), b = 2

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

a × b = 0 → a = 0 or b = 0

So x = −2 or x = 2.

x = −2 or x = 2

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 −2 and 2 cut the line into 3 parts.

Test one number in each part: −3, −1, 3.

f′(−3) = 15, f′(−1) = −9, f′(3) = 15

f′(−3) is 15: positive.

So f goes up there.

f′(−1) is −9: negative.

So f goes down there.

f′(3) is 15: positive.

So f goes up there.

4 Max and min

At x = −2, f goes up, then down.

That is a maximum.

max

At x = 2, f goes down, then up.

That is a minimum.

min

Its y: put −2 into f.

We get 16.

f(−2) = (−2)³ − 12(−2) = −8 + 24 = 16

Its y: put 2 into f.

We get −16.

f(2) = 2³ − 12(2) = 8 − 24 = −16

So: a maximum at (−2, 16), and a minimum at (2, −16).

= max (−2, 16), min (2, −16)

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