Steply

Turning points of f(x) = x^2 - 4x + 1

Step by step, with the reason for every step · Maximum, minimum and optimization · Calculus

Step by step

The question

Question: find the maximum and minimum points of f(x) = x² − 4x + 1.

f(x) = x2 − 4x + 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) = 2x^(2 − 1) − 4 × 1 + 0

Now we have f′(x) = 2x − 4.

= 2x − 4

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
2x − 4 = 0

Take out the 2.

2(x − 2) = 0

So x = 2.

x = 2

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 2 cut the line into 2 parts.

Test one number in each part: 1, 3.

f′(1) = −2, f′(3) = 2

f′(1) is −2: negative.

So f goes down there.

f′(3) is 2: positive.

So f goes up there.

0

4 Max and min

At x = 2, f goes down, then up.

That is a minimum.

min

Its y: put 2 into f.

We get −3.

f(2) = 22 − 4(2) + 1 = 4 − 8 + 1 = −3

So: a minimum at (2, −3).

= min (2, −3)

5 Check

Let's check with the graph.

Down, then up: 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
= min (2, −3)

Your turn

Try this one yourself, then check:

Find the maximum and minimum points of f(x) = x2 + 4x − 2
Show the answer
min (−2, −6)

More from Maximum, minimum and optimization

The whole lesson: Maximum, minimum and optimization →