Absolute max and min of f(x) = x^3 - 3x + 1 on [-2, 0] Step by Step

Step by step

The question

Question: find the absolute maximum and minimum of f(x) = x³ − 3x + 1 on [−2, 0].

f(x) = x³ − 3x + 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 × 1 + 0

Now we have f′(x) = 3x² − 3.

= 3x² − 3

Critical points: where the slope is 0.

Solve f′(x) = 0.

f′(x) = 0

3x² − 3 = 0

Take out the 3.

3(x² − 1) = 0

x² − 1 is a difference of two squares: x² minus 1 squared.

So it is (x − 1)(x + 1).

3(x + 1)(x − 1) = 0

a² − b² = (a − b)(a + b), b = 1

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 = 1.

x = −1 or x = 1

2 Keep the ones inside

x = −1 is inside the interval.

Keep it.

x = −1 in [−2, 0] ✓

x = 1 is outside the interval.

Throw it away.

x = 1 not in [−2, 0] ✗

3 Test every candidate

The left end: f of −2 is −1.

f(−2) = (−2)³ − 3(−2) + 1 = −8 + 6 + 1 = −1

The critical point: f of −1 is 3.

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

The right end: f of 0 is 1.

f(0) = 0³ − 3(0) + 1 = 0 + 0 + 1 = 1

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 3, at x = −1.

That is the absolute maximum.

max : 3 at x = −1

The smallest value is −1, at x = −2.

That is the absolute minimum.

min : −1 at x = −2

5 Check

Let's check with the graph, only from −2 to 0.

The highest point is at x = −1, and the lowest at x = −2.

Right!

Let's look back.

First: critical points.

Then: keep the ones inside.

Then: test every candidate.

Then: biggest and smallest.

Last: check.

Answer