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

Step by step

The question

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

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 [−2, 2] ✓

3 Test every candidate

The left end: f of −2 is −5.

f(−2) = −(−2)² + 2(−2) + 3 = −4 − 4 + 3 = −5

The critical point: f of 1 is 4.

f(1) = −1² + 2(1) + 3 = −1 + 2 + 3 = 4

The right end: f of 2 is 3.

f(2) = −2² + 2(2) + 3 = −4 + 4 + 3 = 3

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 = 1.

That is the absolute maximum.

max : 4 at x = 1

The smallest value is −5, at x = −2.

That is the absolute minimum.

min : −5 at x = −2

5 Check

Let's check with the graph, only from −2 to 2.

The highest point is at x = 1, 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