Absolute maximum and minimum of f(x) = x^2 - 4x + 1 on [0, 5] Step by Step

Step by step

The question

Question: find the absolute maximum and minimum of f(x) = x² − 4x + 1 on [0, 5].

f(x) = x² − 4x + 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) = 2x^(2 − 1) − 4 × 1 + 0

Now we have f′(x) = 2x − 4.

= 2x − 4

Critical points: where the slope is 0.

Solve f′(x) = 0.

f′(x) = 0

2x − 4 = 0

Take out the 2.

2(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, 5] ✓

3 Test every candidate

The left end: f of 0 is 1.

f(0) = 0² − 4(0) + 1 = 0 + 0 + 1 = 1

The critical point: f of 2 is −3.

f(2) = 2² − 4(2) + 1 = 4 − 8 + 1 = −3

The right end: f of 5 is 6.

f(5) = 5² − 4(5) + 1 = 25 − 20 + 1 = 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 6, at x = 5.

That is the absolute maximum.

max : 6 at x = 5

The smallest value is −3, at x = 2.

That is the absolute minimum.

min : −3 at x = 2

5 Check

Let's check with the graph, only from 0 to 5.

The highest point is at x = 5, 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