Step by step
The question
Question: find the maximum and minimum points of f(x) = x³.
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.
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.
3x² is 0 only when x is 0.
So x = 0.
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 0 cut the line into 2 parts.
Test one number in each part: −1, 1.
f′(−1) is 3: positive.
So f goes up there.
f′(1) is 3: positive.
So f goes up there.
4 Max and min
At x = 0, f does not turn: it keeps going the same way.
Not a max, not a min.
The graph never turns.
So there is no maximum or minimum point.
5 Check
Let's check with the graph.
Up, 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.