Absolute maximum and minimum of f(x) = x^2 - 8x + 10 on [1, 6] Step by Step

Step by step

The question

Question: find the absolute maximum and minimum of f(x) = x² − 8x + 10 on [1, 6].

f(x) = x² − 8x + 10

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) − 8 × 1 + 0

Now we have f′(x) = 2x − 8.

= 2x − 8

Critical points: where the slope is 0.

Solve f′(x) = 0.

f′(x) = 0

2x − 8 = 0

Take out the 2.

2(x − 4) = 0

So x = 4.

x = 4

2 Keep the ones inside

x = 4 is inside the interval.

Keep it.

x = 4 in [1, 6] ✓

3 Test every candidate

The left end: f of 1 is 3.

f(1) = 1² − 8(1) + 10 = 1 − 8 + 10 = 3

The critical point: f of 4 is −6.

f(4) = 4² − 8(4) + 10 = 16 − 32 + 10 = −6

The right end: f of 6 is −2.

f(6) = 6² − 8(6) + 10 = 36 − 48 + 10 = −2

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 3, at x = 1.

That is the absolute maximum.

max : 3 at x = 1

The smallest value is −6, at x = 4.

That is the absolute minimum.

min : −6 at x = 4

5 Check

Let's check with the graph, only from 1 to 6.

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

Right!

Let's look back.

First: critical points.

Then: keep the ones inside.

Then: test every candidate.

Then: biggest and smallest.

Last: check.

Answer