Absolute max and min of f(x) = x^2 - 6x + 5 on [0, 4] Step by Step

Step by step

The question

Question: find the absolute maximum and minimum of f(x) = x² − 6x + 5 on [0, 4].

f(x) = x² − 6x + 5

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) − 6 × 1 + 0

Now we have f′(x) = 2x − 6.

= 2x − 6

Critical points: where the slope is 0.

Solve f′(x) = 0.

f′(x) = 0

2x − 6 = 0

Take out the 2.

2(x − 3) = 0

So x = 3.

x = 3

2 Keep the ones inside

x = 3 is inside the interval.

Keep it.

x = 3 in [0, 4] ✓

3 Test every candidate

The left end: f of 0 is 5.

f(0) = 0² − 6(0) + 5 = 0 + 0 + 5 = 5

The critical point: f of 3 is −4.

f(3) = 3² − 6(3) + 5 = 9 − 18 + 5 = −4

The right end: f of 4 is −3.

f(4) = 4² − 6(4) + 5 = 16 − 24 + 5 = −3

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 5, at x = 0.

That is the absolute maximum.

max : 5 at x = 0

The smallest value is −4, at x = 3.

That is the absolute minimum.

min : −4 at x = 3

5 Check

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

The highest point is at x = 0, and the lowest at x = 3.

Right!

Let's look back.

First: critical points.

Then: keep the ones inside.

Then: test every candidate.

Then: biggest and smallest.

Last: check.

Answer