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Where is f(x) = x^3 - 3x increasing

Step by step, with the reason for every step · Maximum, minimum and optimization · Calculus

Step by step

The question

Question: find where f(x) = x³ − 3x is increasing and where it is decreasing.

f(x) = x3 − 3x

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 × 1

Now we have f′(x) = 3x² − 3.

= 3x2 − 3

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 − 3 = 0

Take out the 3.

3(x2 − 1) = 0

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

So it is (x − 1)(x + 1).

3(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) = 9, f′(0) = −3, f′(2) = 9

f′(−2) is 9: positive.

So f goes up there.

f′(0) is −3: negative.

So f goes down there.

f′(2) is 9: positive.

So f goes up there.

0
0

4 Read the table

f is increasing, going up, where f′ is positive: x < −1 and x > 1.

up ↗ : x < −1 and x > 1

f is decreasing, going down, where f′ is negative: −1 < x < 1.

down ↘ : −1 < x < 1

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: read the table.

Last: check.

Answer
up ↗ : x < −1 and x > 1

Your turn

Try this one yourself, then check:

Find where f(x) = x3 + 3x2 + 3
Show the answer
increasing: x < −2 and x > 0; decreasing: −2 < x < 0

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