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

Step by step

The question

Question: find the maximum and minimum points of f(x) = x² − 6x + 1 with the second derivative test.

f(x) = x² − 6x + 1

1 Find f′(x)

Use the power rule on each term.

f′(x) = 2x^(2 − 1) − 6 × 1 + 0

Now we have f′(x) = 2x − 6.

= 2x − 6

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 − 6 = 0

Take out the 2.

2(x − 3) = 0

So x = 3.

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) = 2 × 1 + 0

Now we have f″(x) = 2.

= 2

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 2: positive.

A cup.

So a minimum.

f″(3) = 2 > 0 cup min

Careful!

Positive f″ is a cup.

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

Careful: not f″(3) > 0 → max ✗

Its y: put 3 into f.

We get −8.

f(3) = 3² − 6(3) + 1 = 9 − 18 + 1 = −8

So: a minimum at (3, −8).

= min (3, −8)

5 Check

Let's check with the graph.

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