Worked examples
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.
▶ Watch it solved
The question
Question: find the maximum and minimum points of f(x) = x³ − 3x with the second derivative test.
f(x) = x3 − 3x
1 Find f′(x)
Use the power rule on each term.
f′(x) = 3x^(3 − 1) − 3 × 1
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: negative.
A cap.
So a maximum.
f″(−1) = 6(−1) = −6 < 0 cap max
Careful!
Negative f″ is a cap.
The top of a cap is a maximum, not a minimum.
Careful: not f″(−1) < 0 → min ✗
At x = 1, f″ is 6: positive.
A cup.
So a minimum.
f″(1) = 6(1) = 6 > 0 cup min
Its y: put −1 into f.
We get 2.
f(−1) = (−1)3 − 3(−1) = −1 + 3 = 2
Its y: put 1 into f.
We get −2.
f(1) = 13 − 3(1) = 1 − 3 = −2
So: a maximum at (−1, 2), and a minimum at (1, −2).
= max (−1, 2), min (1, −2)
5 Check
Let's check with the graph.
A cap at the maximum, and 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 = max (−1, 2), min (1, −2)
▶ Watch it solved
The question
Question: find the maximum and minimum points of f(x) = x³ − 3x² − 9x + 2 with the second derivative test.
f(x) = x3 − 3x2 − 9x + 2
1 Find f′(x)
Use the power rule on each term.
f′(x) = 3x^(3 − 1) − 3 × 2x^(2 − 1) − 9 × 1 + 0
Now we have f′(x) = 3x² − 6x − 9.
= 3x2 − 6x − 9
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 − 6x − 9 = 0
Take out the 3.
3(x2 − 2x − 3) = 0
We need two numbers that multiply to −3 and add to −2: 1 and −3.
3(x + 1)(x − 3) = 0
1 × (−3) = −3, 1 + (−3) = −2
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 = 3.
x = −1 or 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) = 3 × 2x^(2 − 1) − 6 × 1 + 0
Now we have f″(x) = 6x − 6.
= 6x − 6
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 −12: negative.
A cap.
So a maximum.
f″(−1) = 6(−1) − 6 = −6 − 6 = −12 < 0 cap max
Careful!
Negative f″ is a cap.
The top of a cap is a maximum, not a minimum.
Careful: not f″(−1) < 0 → min ✗
At x = 3, f″ is 12: positive.
A cup.
So a minimum.
f″(3) = 6(3) − 6 = 18 − 6 = 12 > 0 cup min
Its y: put −1 into f.
We get 7.
f(−1) = (−1)3 − 3(−1)2 − 9(−1) + 2 = −1 − 3 + 9 + 2 = 7
Its y: put 3 into f.
We get −25.
f(3) = 33 − 3(3)2 − 9(3) + 2 = 27 − 27 − 27 + 2 = −25
So: a maximum at (−1, 7), and a minimum at (3, −25).
= max (−1, 7), min (3, −25)
5 Check
Let's check with the graph.
A cap at the maximum, and 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 = max (−1, 7), min (3, −25)
▶ Watch it solved
