Absolute max and min of f(x) = x^3 - 6x^2 + 9x on [0, 4] Step by Step

Step by step

The question

Question: find the absolute maximum and minimum of f(x) = x³ − 6x² + 9x on [0, 4].

f(x) = x³ − 6x² + 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) − 6 × 2x^(2 − 1) + 9 × 1

Now we have f′(x) = 3x² − 12x + 9.

= 3x² − 12x + 9

Critical points: where the slope is 0.

Solve f′(x) = 0.

f′(x) = 0

3x² − 12x + 9 = 0

Take out the 3.

3(x² − 4x + 3) = 0

We need two numbers that multiply to 3 and add to −4: −1 and −3.

3(x − 1)(x − 3) = 0

−1 × (−3) = 3, −1 + (−3) = −4

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 [0, 4] ✓

x = 3 is inside the interval.

Keep it.

x = 3 in [0, 4] ✓

3 Test every candidate

The left end: f of 0 is 0.

f(0) = 0³ − 6(0)² + 9(0) = 0 + 0 + 0 = 0

The critical point: f of 1 is 4.

f(1) = 1³ − 6(1)² + 9(1) = 1 − 6 + 9 = 4

The critical point: f of 3 is 0.

f(3) = 3³ − 6(3)² + 9(3) = 27 − 54 + 27 = 0

The right end: f of 4 is 4.

f(4) = 4³ − 6(4)² + 9(4) = 64 − 96 + 36 = 4

4 Biggest and smallest

Careful!

The minimum is at an end, where the slope is not 0.

That is why we test the ends.

Careful: the ends count too

The biggest value is 4, at x = 1 and x = 4.

That is the absolute maximum.

max : 4 at x = 1 and x = 4

The smallest value is 0, at x = 0 and x = 3.

That is the absolute minimum.

min : 0 at x = 0 and x = 3

5 Check

Let's check with the graph, only from 0 to 4.

The highest point is at x = 1 and x = 4, and the lowest at x = 0 and x = 3.

Right!

Let's look back.

First: critical points.

Then: keep the ones inside.

Then: test every candidate.

Then: biggest and smallest.

Last: check.

Answer