Absolute max and min of f(x) = x^3 - 27x on [0, 4] Step by Step

Step by step

The question

Question: find the absolute maximum and minimum of f(x) = x³ − 27x on [0, 4].

f(x) = x³ − 27x

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) − 27 × 1

Now we have f′(x) = 3x² − 27.

= 3x² − 27

Critical points: where the slope is 0.

Solve f′(x) = 0.

f′(x) = 0

3x² − 27 = 0

Take out the 3.

3(x² − 9) = 0

x² − 9 is a difference of two squares: x² minus 3 squared.

So it is (x − 3)(x + 3).

3(x + 3)(x − 3) = 0

a² − b² = (a − b)(a + b), b = 3

A product is 0 only when one of its parts is 0.

a × b = 0 → a = 0 or b = 0

So x = −3 or x = 3.

x = −3 or x = 3

2 Keep the ones inside

x = −3 is outside the interval.

Throw it away.

x = −3 not 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 − 27(0) = 0 + 0 = 0

The critical point: f of 3 is −54.

f(3) = 3³ − 27(3) = 27 − 81 = −54

The right end: f of 4 is −44.

f(4) = 4³ − 27(4) = 64 − 108 = −44

4 Biggest and smallest

Careful!

The maximum 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 0, at x = 0.

That is the absolute maximum.

max : 0 at x = 0

The smallest value is −54, at x = 3.

That is the absolute minimum.

min : −54 at x = 3

5 Check

Let's check one more point: x = 2 gives −46.

It is between the min and the max.

Right!

f(2) = −46: between −54 and 0

Let's look back.

First: critical points.

Then: keep the ones inside.

Then: test every candidate.

Then: biggest and smallest.

Last: check.

Answer