Turning points of 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.

f(x) = −x³ + 3x + 1

The derivative is the slope.

Where it is positive, the graph goes up.

Where it is negative, the graph goes down.

f′(x) > 0: going up ↗

f′(x) < 0: going down ↘

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

Careful!

f′ = 0 alone does not tell max or min.

The signs on each side tell us.

Careful: f′(x) = 0 is not enough: look at the signs

3 Sign table

A sign table.

The numbers −1 and 1 cut the line into 3 parts.

Test one number in each part: −2, 0, 2.

f′(−2) = −9, f′(0) = 3, f′(2) = −9

f′(−2) is −9: negative.

So f goes down there.

f′(0) is 3: positive.

So f goes up there.

f′(2) is −9: negative.

So f goes down there.

4 Max and min

At x = −1, f goes down, then up.

That is a minimum.

min

At x = 1, f goes up, then down.

That is a maximum.

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.

Down, then up, then down: just like the table.

Right!

Let's look back.

First: find f′(x).

Then: solve f′(x) = 0.

Then: sign table.

Then: max and min.

Last: check.

Answer