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

Step by step

The question

Question: find the maximum and minimum points of f(x) = x³ + 3x² − 9x with the second derivative test.

f(x) = x³ + 3x² − 9x

1 Find f′(x)

Use the power rule on each term.

f′(x) = 3x^(3 − 1) + 3 × 2x^(2 − 1) − 9 × 1

Now we have f′(x) = 3x² + 6x − 9.

= 3x² + 6x − 9

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 + 6x − 9 = 0

Take out the 3.

3(x² + 2x − 3) = 0

We need two numbers that multiply to −3 and add to 2: 3 and −1.

3(x + 3)(x − 1) = 0

3 × (−1) = −3, 3 + (−1) = 2

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

a × b = 0 → a = 0 or b = 0

So x = −3 or x = 1.

x = −3 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) = 3 × 2x^(2 − 1) + 6 × 1 + 0

Now we have f″(x) = 6x + 6.

= 6x + 6

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 = −3, f″ is −12: negative.

A cap.

So a maximum.

f″(−3) = 6(−3) + 6 = −18 + 6 = −12 < 0 cap max

Careful!

Negative f″ is a cap.

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

Careful: not f″(−3) < 0 → min ✗

At x = 1, f″ is 12: positive.

A cup.

So a minimum.

f″(1) = 6(1) + 6 = 6 + 6 = 12 > 0 cup min

Its y: put −3 into f.

We get 27.

f(−3) = (−3)³ + 3(−3)² − 9(−3) = −27 + 27 + 27 = 27

Its y: put 1 into f.

We get −5.

f(1) = 1³ + 3(1)² − 9(1) = 1 + 3 − 9 = −5

So: a maximum at (−3, 27), and a minimum at (1, −5).

= max (−3, 27), min (1, −5)

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