Concavity and inflection points of f(x) = x^3 + 3x^2

Step by step

The question

Question: find the concavity and the inflection points of f(x) = x³ + 3x² − 2.

f(x) = x³ + 3x² − 2

Concave up means the curve bends up, like a cup.

That is where f″ is positive.

Concave down bends down, like a cap: f″ is negative.

cup concave up: f″ > 0

cap concave down: f″ < 0

1 Find f″(x)

First f′: the power rule on each term.

f′(x) = 3x^(3 − 1) + 3 × 2x^(2 − 1) + 0

Now we have f′(x) = 3x² + 6x.

= 3x² + 6x

Then f″: the power rule again, on f′.

f″(x) = 3 × 2x^(2 − 1) + 6 × 1

Now we have f″(x) = 6x + 6.

= 6x + 6

2 Solve f″(x) = 0

The bending can only change where f″ is 0.

So we solve f″(x) = 0.

f″(x) = 0

6x + 6 = 0

Take out the 6.

6(x + 1) = 0

So x = −1.

x = −1

3 Sign table for f″

A sign table for f″.

The number −1 cuts the line into 2 parts.

Test one number in each part: −2, 0.

f″(−2) = −6, f″(0) = 6

f″(−2) is −6: negative.

So f bends down there: a cap.

cap

f″(0) is 6: positive.

So f bends up there: a cup.

cup

Careful!

Concave up is about the bending, not about going up.

A cup can go down on one side.

Careful: concave up is not "going up"

4 Inflection points

At x = −1, the bending changes: an inflection point.

infl.

Its y: put −1 into f.

We get 0.

f(−1) = (−1)³ + 3(−1)² − 2 = −1 + 3 − 2 = 0

f is concave up where f″ is positive: x > −1.

up ∪ : x > −1

f is concave down where f″ is negative: x < −1.

down ∩ : x < −1

And the inflection point is (−1, 0).

inflection : (−1, 0)

5 Check

Let's check with the graph.

A cap, then a cup.

It changes at the inflection point.

Right!

Let's look back.

First: find f″(x).

Then: solve f″(x) = 0.

Then: sign table for f″.

Then: inflection points.

Last: check.

Answer