Absolute maximum and minimum of f(x) = x^3 - 3x^2

Step by step

The question

Question: find the absolute maximum and minimum of f(x) = x³ − 3x² − 9x on [−2, 4].

f(x) = x³ − 3x² − 9x

On a closed interval, the highest and lowest values are at an end, or where the slope is 0.

So we test those, and only those.

candidates: the two ends + where f′ = 0

1 Critical points

Use the power rule on each term.

f′(x) = 3x^(3 − 1) − 3 × 2x^(2 − 1) − 9 × 1

Now we have f′(x) = 3x² − 6x − 9.

= 3x² − 6x − 9

Critical points: where the slope is 0.

Solve f′(x) = 0.

f′(x) = 0

3x² − 6x − 9 = 0

Take out the 3.

3(x² − 2x − 3) = 0

We need two numbers that multiply to −3 and add to −2: 1 and −3.

3(x + 1)(x − 3) = 0

1 × (−3) = −3, 1 + (−3) = −2

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 = 3.

x = −1 or x = 3

2 Keep the ones inside

x = −1 is inside the interval.

Keep it.

x = −1 in [−2, 4] ✓

x = 3 is inside the interval.

Keep it.

x = 3 in [−2, 4] ✓

3 Test every candidate

The left end: f of −2 is −2.

f(−2) = (−2)³ − 3(−2)² − 9(−2) = −8 − 12 + 18 = −2

The critical point: f of −1 is 5.

f(−1) = (−1)³ − 3(−1)² − 9(−1) = −1 − 3 + 9 = 5

The critical point: f of 3 is −27.

f(3) = 3³ − 3(3)² − 9(3) = 27 − 27 − 27 = −27

The right end: f of 4 is −20.

f(4) = 4³ − 3(4)² − 9(4) = 64 − 48 − 36 = −20

4 Biggest and smallest

The biggest value is 5, at x = −1.

That is the absolute maximum.

max : 5 at x = −1

The smallest value is −27, at x = 3.

That is the absolute minimum.

min : −27 at x = 3

5 Check

Let's check one more point: x = 1 gives −11.

It is between the min and the max.

Right!

f(1) = −11: between −27 and 5

Let's look back.

First: critical points.

Then: keep the ones inside.

Then: test every candidate.

Then: biggest and smallest.

Last: check.

Answer