Absolute max and min of f(x) = 2x^2 - 8x on [0, 3] Step by Step

Step by step

The question

Question: find the absolute maximum and minimum of f(x) = 2x² − 8x on [0, 3].

f(x) = 2x² − 8x

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 × 2x^(2 − 1) − 8 × 1

Now we have f′(x) = 4x − 8.

= 4x − 8

Critical points: where the slope is 0.

Solve f′(x) = 0.

f′(x) = 0

4x − 8 = 0

Take out the 4.

4(x − 2) = 0

So x = 2.

x = 2

2 Keep the ones inside

x = 2 is inside the interval.

Keep it.

x = 2 in [0, 3] ✓

3 Test every candidate

The left end: f of 0 is 0.

f(0) = 2(0)² − 8(0) = 0 + 0 = 0

The critical point: f of 2 is −8.

f(2) = 2(2)² − 8(2) = 8 − 16 = −8

The right end: f of 3 is −6.

f(3) = 2(3)² − 8(3) = 18 − 24 = −6

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 −8, at x = 2.

That is the absolute minimum.

min : −8 at x = 2

5 Check

Let's check with the graph, only from 0 to 3.

The highest point is at x = 0, 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