Absolute max and min of f(x) = 2x^3 - 3x^2 - 12x on [-2, 3] Step by Step

Step by step

The question

Question: find the absolute maximum and minimum of f(x) = 2x³ − 3x² − 12x on [−2, 3].

f(x) = 2x³ − 3x² − 12x

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) = 2 × 3x^(3 − 1) − 3 × 2x^(2 − 1) − 12 × 1

Now we have f′(x) = 6x² − 6x − 12.

= 6x² − 6x − 12

Critical points: where the slope is 0.

Solve f′(x) = 0.

f′(x) = 0

6x² − 6x − 12 = 0

Take out the 6.

6(x² − x − 2) = 0

We need two numbers that multiply to −2 and add to −1: 1 and −2.

6(x + 1)(x − 2) = 0

1 × (−2) = −2, 1 + (−2) = −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 = 2.

x = −1 or x = 2

2 Keep the ones inside

x = −1 is inside the interval.

Keep it.

x = −1 in [−2, 3] ✓

x = 2 is inside the interval.

Keep it.

x = 2 in [−2, 3] ✓

3 Test every candidate

The left end: f of −2 is −4.

f(−2) = 2(−2)³ − 3(−2)² − 12(−2) = −16 − 12 + 24 = −4

The critical point: f of −1 is 7.

f(−1) = 2(−1)³ − 3(−1)² − 12(−1) = −2 − 3 + 12 = 7

The critical point: f of 2 is −20.

f(2) = 2(2)³ − 3(2)² − 12(2) = 16 − 12 − 24 = −20

The right end: f of 3 is −9.

f(3) = 2(3)³ − 3(3)² − 12(3) = 54 − 27 − 36 = −9

4 Biggest and smallest

The biggest value is 7, at x = −1.

That is the absolute maximum.

max : 7 at x = −1

The smallest value is −20, at x = 2.

That is the absolute minimum.

min : −20 at x = 2

5 Check

Let's check one more point: x = 0.5 gives −6.5.

It is between the min and the max.

Right!

f(0.5) = −6.5: between −20 and 7

Let's look back.

First: critical points.

Then: keep the ones inside.

Then: test every candidate.

Then: biggest and smallest.

Last: check.

Answer