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Inflection points of f(x) = x^4 - 2x^3

Step by step, with the reason for every step · Concavity and inflection points · Calculus 1

Step by step

The question

Question: find the concavity and the inflection points of f(x) = x⁴ − 2x³.

f(x) = x4 − 2x3

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) = 4x^(4 − 1) − 2 × 3x^(3 − 1)

Now we have f′(x) = 4x³ − 6x².

= 4x3 − 6x2

Then f″: the power rule again, on f′.

f″(x) = 4 × 3x^(3 − 1) − 6 × 2x^(2 − 1)

Now we have f″(x) = 12x² − 12x.

= 12x2 − 12x

2 Solve f″(x) = 0

The bending can only change where f″ is 0.

So we solve f″(x) = 0.

f″(x) = 0
12x2 − 12x = 0

Every term has 12x: take it out.

12x(x − 1) = 0

A product is 0 only when one of its parts is 0.

a × b = 0 → a = 0 or b = 0

So x = 0 or x = 1.

x = 0 or x = 1

3 Sign table for f″

A sign table for f″.

The numbers 0 and 1 cut the line into 3 parts.

Test one number in each part: −1, 0.5, 2.

f″(−1) = 24, f″(0.5) = −3, f″(2) = 24

f″(−1) is 24: positive.

So f bends up there: a cup.

cup

f″(0.5) is −3: negative.

So f bends down there: a cap.

cap

f″(2) is 24: positive.

So f bends up there: a cup.

cup
0
0

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 = 0, the bending changes: an inflection point.

infl.

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

infl.

Its y: put 0 into f.

We get 0.

f(0) = 04 − 2(0)3 = 0 + 0 = 0

Its y: put 1 into f.

We get −1.

f(1) = 14 − 2(1)3 = 1 − 2 = −1

f is concave up where f″ is positive: x < 0 and x > 1.

up ∪ : x < 0 and x > 1

f is concave down where f″ is negative: 0 < x < 1.

down ∩ : 0 < x < 1

And the inflection points are (0, 0) and (1, −1).

inflection : (0, 0), (1, −1)

5 Check

Let's check with the graph.

A cup, then a cap, then a cup.

It changes at the inflection points.

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
inflection : (0, 0), (1, −1)

Your turn

Try this one yourself, then check:

Find the concavity and the inflection points of f(x) = x4 − 6x2 + 5
Show the answer
concave up: x < −1 and x > 1; concave down: −1 < x < 1; inflection: (−1, 0), (1, 0)

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