Use the second derivative test on f(x) = 2x^3

Step by step

The question

Question: find the maximum and minimum points of f(x) = 2x³ − 6x with the second derivative test.

f(x) = 2x³ − 6x

1 Find f′(x)

Use the power rule on each term.

f′(x) = 2 × 3x^(3 − 1) − 6 × 1

Now we have f′(x) = 6x² − 6.

= 6x² − 6

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

6x² − 6 = 0

Take out the 6.

6(x² − 1) = 0

x² − 1 is a difference of two squares: x² minus 1 squared.

So it is (x − 1)(x + 1).

6(x + 1)(x − 1) = 0

a² − b² = (a − b)(a + b), b = 1

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

a × b = 0 → a = 0 or b = 0

So x = −1 or x = 1.

x = −1 or x = 1

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) = 6 × 2x^(2 − 1) + 0

Now we have f″(x) = 12x.

= 12x

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 = −1, f″ is −12: negative.

A cap.

So a maximum.

f″(−1) = 12(−1) = −12 < 0 cap max

Careful!

Negative f″ is a cap.

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

Careful: not f″(−1) < 0 → min ✗

At x = 1, f″ is 12: positive.

A cup.

So a minimum.

f″(1) = 12(1) = 12 > 0 cup min

Its y: put −1 into f.

We get 4.

f(−1) = 2(−1)³ − 6(−1) = −2 + 6 = 4

Its y: put 1 into f.

We get −4.

f(1) = 2(1)³ − 6(1) = 2 − 6 = −4

So: a maximum at (−1, 4), and a minimum at (1, −4).

= max (−1, 4), min (1, −4)

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