Use the second derivative test on f(x) = -x^3 + 3x + 1 Step by Step

Step by step

The question

Question: find the maximum and minimum points of f(x) = −x³ + 3x + 1 with the second derivative test.

f(x) = −x³ + 3x + 1

1 Find f′(x)

Use the power rule on each term.

f′(x) = −3x^(3 − 1) + 3 × 1 + 0

Now we have f′(x) = −3x² + 3.

= −3x² + 3

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 + 3 = 0

Take out the −3.

−3(x² − 1) = 0

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

So it is (x − 1)(x + 1).

−3(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) = −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 = −1, f″ is 6: positive.

A cup.

So a minimum.

f″(−1) = −6(−1) = 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 = 1, f″ is −6: negative.

A cap.

So a maximum.

f″(1) = −6(1) = −6 < 0 cap max

Its y: put −1 into f.

We get −1.

f(−1) = −(−1)³ + 3(−1) + 1 = 1 − 3 + 1 = −1

Its y: put 1 into f.

We get 3.

f(1) = −1³ + 3(1) + 1 = −1 + 3 + 1 = 3

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

= min (−1, −1), max (1, 3)

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