Step by step
The question
Question: find the absolute maximum and minimum of f(x) = x³ + 3x² − 9x − 4 on [−4, 2].
On a closed interval, the highest and lowest values are at an end, or inside.
Inside, a top or a bottom is where the curve turns.
There it is flat: the slope is 0.
So we test the ends and those points.
1 Critical points
Use the power rule on each term.
Now we have f′(x) = 3x² + 6x − 9.
Critical points: where the slope is 0.
Solve f′(x) = 0.
That is 3x² + 6x − 9 = 0.
Take out the 3.
To factor x² + 2x − 3, the rule: we need two numbers that multiply to −3 and add to 2.
They are 3 and −1: they multiply to −3 and add to 2.
So we write it as 3(x + 3)(x − 1).
A product is 0 only when one of its parts is 0.
So x = −3 or x = 1.
2 Keep the ones inside
x = −3 is inside the interval.
Keep it.
x = 1 is inside the interval.
Keep it.
3 Test every candidate
The left end: f of −4 is 16.
The critical point: f of −3 is 23.
The critical point: f of 1 is −9.
The right end: f of 2 is −2.
4 Biggest and smallest
The biggest value is 23, at x = −3.
That is the absolute maximum.
The smallest value is −9, at x = 1.
That is the absolute minimum.
5 Check
Let's check one more point.
Every x inside must give a value between the min and the max.
Take the middle, x = −1: it gives 7.
Between them.
Right!
Let's look back.
First: critical points.
Then: keep the ones inside.
Then: test every candidate.
Then: biggest and smallest.
Last: check.