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

Step by step

The question

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

f(x) = −x² + 4x

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

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 0.

f(0) = −(0)² + 4(0) = 0 + 0 = 0

The critical point: f of 2 is 4.

f(2) = −2² + 4(2) = −4 + 8 = 4

The right end: f of 5 is −5.

f(5) = −5² + 4(5) = −25 + 20 = −5

4 Biggest and smallest

Careful!

The minimum 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 4, at x = 2.

That is the absolute maximum.

max : 4 at x = 2

The smallest value is −5, at x = 5.

That is the absolute minimum.

min : −5 at x = 5

5 Check

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

The highest point is at x = 2, and the lowest at x = 5.

Right!

Let's look back.

First: critical points.

Then: keep the ones inside.

Then: test every candidate.

Then: biggest and smallest.

Last: check.

Answer