Use the second derivative test on f(x) = x^3

Step by step

The question

Question: find the maximum and minimum points of f(x) = x³ − 12x with the second derivative test.

f(x) = x³ − 12x

1 Find f′(x)

Use the power rule on each term.

f′(x) = 3x^(3 − 1) − 12 × 1

Now we have f′(x) = 3x² − 12.

= 3x² − 12

2 Solve f′(x) = 0

At a turning point the graph is flat for a moment: the slope is 0.

So we solve f′(x) = 0.

f′(x) = 0

3x² − 12 = 0

Take out the 3.

3(x² − 4) = 0

x² − 4 is a difference of two squares: x² minus 2 squared.

So it is (x − 2)(x + 2).

3(x + 2)(x − 2) = 0

a² − b² = (a − b)(a + b), b = 2

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

a × b = 0 → a = 0 or b = 0

So x = −2 or x = 2.

x = −2 or x = 2

3 Find f″(x)

f double prime is the derivative of f′.

It tells how the slope changes.

f″(x): the derivative of f′(x)

Use the power rule again, now on f′.

f″(x) = 3 × 2x^(2 − 1) + 0

Now we have f″(x) = 6x.

= 6x

If f″ is positive, the slope is growing: the curve bends up, like a cup.

Its bottom is a minimum.

If f″ is negative, it bends down, like a cap.

Its top is a maximum.

cup f″ > 0: a cup, a min

cap f″ < 0: a cap, a max

4 Test each point

At x = −2, f″ is −12: negative.

A cap.

So a maximum.

f″(−2) = 6(−2) = −12 < 0 cap max

Careful!

Negative f″ is a cap.

The top of a cap is a maximum, not a minimum.

Careful: not f″(−2) < 0 → min ✗

At x = 2, f″ is 12: positive.

A cup.

So a minimum.

f″(2) = 6(2) = 12 > 0 cup min

Its y: put −2 into f.

We get 16.

f(−2) = (−2)³ − 12(−2) = −8 + 24 = 16

Its y: put 2 into f.

We get −16.

f(2) = 2³ − 12(2) = 8 − 24 = −16

So: a maximum at (−2, 16), and a minimum at (2, −16).

= max (−2, 16), min (2, −16)

5 Check

Let's check with the graph.

A cap at the maximum, and a cup at the minimum.

Right!

Let's look back.

First: find f′(x).

Then: solve f′(x) = 0.

Then: find f″(x).

Then: test each point.

Last: check.

Answer