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

Step by step

The question

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

f(x) = −x³ + 3x

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

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 outside the interval.

Throw it away.

x = −1 not in [0, 2] ✗

x = 1 is inside the interval.

Keep it.

x = 1 in [0, 2] ✓

3 Test every candidate

The left end: f of 0 is 0.

f(0) = −(0)³ + 3(0) = 0 + 0 = 0

The critical point: f of 1 is 2.

f(1) = −1³ + 3(1) = −1 + 3 = 2

The right end: f of 2 is −2.

f(2) = −2³ + 3(2) = −8 + 6 = −2

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 2, at x = 1.

That is the absolute maximum.

max : 2 at x = 1

The smallest value is −2, at x = 2.

That is the absolute minimum.

min : −2 at x = 2

5 Check

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

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