Absolute max and min of f(x) = x^3 - 3x^2 + 1 on [-1, 4] Step by Step

Step by step

The question

Question: find the absolute maximum and minimum of f(x) = x³ − 3x² + 1 on [−1, 4].

f(x) = x³ − 3x² + 1

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) + 0

Now we have f′(x) = 3x² − 6x.

= 3x² − 6x

Critical points: where the slope is 0.

Solve f′(x) = 0.

f′(x) = 0

3x² − 6x = 0

Take out the 3.

3(x² − 2x) = 0

We need two numbers that multiply to 0 and add to −2: 0 and −2.

3x(x − 2) = 0

0 × (−2) = 0, 0 + (−2) = −2

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

a × b = 0 → a = 0 or b = 0

So x = 0 or x = 2.

x = 0 or x = 2

2 Keep the ones inside

x = 0 is inside the interval.

Keep it.

x = 0 in [−1, 4] ✓

x = 2 is inside the interval.

Keep it.

x = 2 in [−1, 4] ✓

3 Test every candidate

The left end: f of −1 is −3.

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

The critical point: f of 0 is 1.

f(0) = 0³ − 3(0)² + 1 = 0 + 0 + 1 = 1

The critical point: f of 2 is −3.

f(2) = 2³ − 3(2)² + 1 = 8 − 12 + 1 = −3

The right end: f of 4 is 17.

f(4) = 4³ − 3(4)² + 1 = 64 − 48 + 1 = 17

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 17, at x = 4.

That is the absolute maximum.

max : 17 at x = 4

The smallest value is −3, at x = −1 and x = 2.

That is the absolute minimum.

min : −3 at x = −1 and x = 2

5 Check

Let's check one more point: x = 1.5 gives −2.37.

It is between the min and the max.

Right!

f(1.5) = −2.37: between −3 and 17

Let's look back.

First: critical points.

Then: keep the ones inside.

Then: test every candidate.

Then: biggest and smallest.

Last: check.

Answer