Use the second derivative test on f(x) = -x^3 + 6x^2

Step by step

The question

Question: find the maximum and minimum points of f(x) = −x³ + 6x² − 9x with the second derivative test.

f(x) = −x³ + 6x² − 9x

1 Find f′(x)

Use the power rule on each term.

f′(x) = −3x^(3 − 1) + 6 × 2x^(2 − 1) − 9 × 1

Now we have f′(x) = −3x² + 12x − 9.

= −3x² + 12x − 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 + 12x − 9 = 0

Take out the −3.

−3(x² − 4x + 3) = 0

We need two numbers that multiply to 3 and add to −4: −1 and −3.

−3(x − 1)(x − 3) = 0

−1 × (−3) = 3, −1 + (−3) = −4

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 = 3.

x = 1 or x = 3

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) + 12 × 1 + 0

Now we have f″(x) = −6x + 12.

= −6x + 12

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 6: positive.

A cup.

So a minimum.

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

Careful!

Positive f″ is a cup.

The bottom of a cup is a minimum, not a maximum.

Careful: not f″(1) > 0 → max ✗

At x = 3, f″ is −6: negative.

A cap.

So a maximum.

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

Its y: put 1 into f.

We get −4.

f(1) = −1³ + 6(1)² − 9(1) = −1 + 6 − 9 = −4

Its y: put 3 into f.

We get 0.

f(3) = −3³ + 6(3)² − 9(3) = −27 + 54 − 27 = 0

So: a minimum at (1, −4), and a maximum at (3, 0).

= min (1, −4), max (3, 0)

5 Check

Let's check with the graph.

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

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