Absolute max and min of f(x) = x^2 - 2x + 3 on [0, 4] Step by Step

Step by step

The question

Question: find the absolute maximum and minimum of f(x) = x² − 2x + 3 on [0, 4].

f(x) = x² − 2x + 3

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) − 2 × 1 + 0

Now we have f′(x) = 2x − 2.

= 2x − 2

Critical points: where the slope is 0.

Solve f′(x) = 0.

f′(x) = 0

2x − 2 = 0

Take out the 2.

2(x − 1) = 0

So x = 1.

x = 1

2 Keep the ones inside

x = 1 is inside the interval.

Keep it.

x = 1 in [0, 4] ✓

3 Test every candidate

The left end: f of 0 is 3.

f(0) = 0² − 2(0) + 3 = 0 + 0 + 3 = 3

The critical point: f of 1 is 2.

f(1) = 1² − 2(1) + 3 = 1 − 2 + 3 = 2

The right end: f of 4 is 11.

f(4) = 4² − 2(4) + 3 = 16 − 8 + 3 = 11

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 11, at x = 4.

That is the absolute maximum.

max : 11 at x = 4

The smallest value is 2, at x = 1.

That is the absolute minimum.

min : 2 at x = 1

5 Check

Let's check with the graph, only from 0 to 4.

The highest point is at x = 4, and the lowest at x = 1.

Right!

Let's look back.

First: critical points.

Then: keep the ones inside.

Then: test every candidate.

Then: biggest and smallest.

Last: check.

Answer