Absolute max and min of f(x) = x^2 + 2x on [-3, 1] Step by Step

Step by step

The question

Question: find the absolute maximum and minimum of f(x) = x² + 2x on [−3, 1].

f(x) = x² + 2x

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

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 [−3, 1] ✓

3 Test every candidate

The left end: f of −3 is 3.

f(−3) = (−3)² + 2(−3) = 9 − 6 = 3

The critical point: f of −1 is −1.

f(−1) = (−1)² + 2(−1) = 1 − 2 = −1

The right end: f of 1 is 3.

f(1) = 1² + 2(1) = 1 + 2 = 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 3, at x = −3 and x = 1.

That is the absolute maximum.

max : 3 at x = −3 and x = 1

The smallest value is −1, at x = −1.

That is the absolute minimum.

min : −1 at x = −1

5 Check

Let's check with the graph, only from −3 to 1.

The highest point is at x = −3 and x = 1, 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