Steply

The function f(x) = x^2 - 6x + 2 has a relative minimum. At which…

Step by step, with the reason for every step · AP Calculus AB Practice Exam 2 · AP Calculus AB

Step by step

The question

Question: find the maximum and minimum points of f(x) = x² − 6x + 2 with the second derivative test.

f(x) = x2 − 6x + 2

There are four choices.

First we solve it, the normal way.

Then we pick.

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

That is 2x − 6 = 0.

2x − 6 = 0

To remove the − 6, we add 6 on both sides.

2x = 6

To remove times 2, we divide both sides by 2.

6 divided by 2 is 3.

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 −7.

f(3) = 32 − 6(3) + 2 = 9 − 18 + 2 = −7

So: a minimum at (3, −7).

min (3, −7)

5 Check

Let's check with the graph.

A cup at the minimum.

Right!

6 Pick the choice

Our answer is min (3, −7).

That is choice A.

A) (3, −7)

7 Why not the others

B is (−3, 29).

Wrong: solved 2x − 6 = 0 with the wrong sign.

B) (−3, 29) ✗ solved 2x − 6 = 0 with the wrong sign

C is (3, 2).

Wrong: put x = 3 into the 2 only.

C) (3, 2) ✗ put x = 3 into the 2 only

D is (6, 2).

Wrong: set f(x) = 2, not f′(x) = 0.

D) (6, 2) ✗ set f(x) = 2, not f′(x) = 0

Let's look back.

First: find f′(x).

Then: solve f′(x) = 0.

Then: find f″(x).

Then: test each point.

Then: check.

Then: pick the choice.

Last: why not the others.

Answer
min (3, −7)

Your turn

Try this one yourself, then check your answer:

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

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