Step by step
The question
Question: find the maximum and minimum points of 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.
1 Find f′(x)
Use the power rule on each term.
Now we have f′(x) = 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.
Take out the 3.
x² − 1 is a difference of two squares: x² minus 1 squared.
So it is (x − 1)(x + 1).
A product is 0 only when one of its parts is 0.
So x = −1 or x = 1.
Careful!
f′ = 0 alone does not tell max or min.
The signs on each side tell us.
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) 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.
4 Max and min
At x = −1, f goes up, then down.
That is a maximum.
At x = 1, f goes down, then up.
That is a minimum.
Its y: put −1 into f.
We get 2.
Its y: put 1 into f.
We get −2.
So: a maximum at (−1, 2), and a minimum at (1, −2).
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.