Step by step: way 1, first derivative
The question
Question: find the maximum and minimum points of f(x) = x³ + 3x² − 9x − 4.
f(x) = x3 + 3x2 − 9x − 4
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 × 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
That is 3x² + 6x − 9 = 0.
3x2 + 6x − 9 = 0
Take out the 3.
3(x2 + 2x − 3) = 0
To factor x² + 2x − 3, the rule: we need two numbers that multiply to −3 and add to 2.
They are 3 and −1: they multiply to −3 and add to 2.
? × ? = −3
? + ? = 2
3 × (−1) = −3, 3 + (−1) = 2 ✓
So we write it as 3(x + 3)(x − 1).
3(x + 3)(x − 1) = 0
A product is 0 only when one of its parts is 0.
a × b = 0 → a = 0 or b = 0
So x = −3 or x = 1.
x = −3 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 −3 and 1 cut the line into 3 parts.
Test one number in each part: −4, −2, 2.
Any number in a part gives the same sign, so we pick easy ones.
f′(−4) = 15, f′(−2) = −9, f′(2) = 15
f′(−4) is 15: positive.
So f goes up there.
f′(−2) is −9: negative.
So f goes down there.
f′(2) is 15: positive.
So f goes up there.
0
0
4 Max and min
At x = −3, f goes up, then down.
That is a maximum.
max
At x = 1, f goes down, then up.
That is a minimum.
min
Its y: put −3 into f.
We get 23.
f(−3) = (−3)3 + 3(−3)2 − 9(−3) − 4 = −27 + 27 + 27 − 4 = 23
Its y: put 1 into f.
We get −9.
f(1) = 13 + 3(1)2 − 9(1) − 4 = 1 + 3 − 9 − 4 = −9
So: a maximum at (−3, 23), and a minimum at (1, −9).
max (−3, 23), min (1, −9)
5 Check
Let's check with the graph.
Up, then 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 max (−3, 23), min (1, −9)
Step by step: way 2, second derivative
The question
Question: find the maximum and minimum points of f(x) = x³ + 3x² − 9x − 4 with the second derivative test.
f(x) = x3 + 3x2 − 9x − 4
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
That is 3x² + 6x − 9 = 0.
3x2 + 6x − 9 = 0
Take out the 3.
3(x2 + 2x − 3) = 0
To factor x² + 2x − 3, the rule: we need two numbers that multiply to −3 and add to 2.
They are 3 and −1: they multiply to −3 and add to 2.
? × ? = −3
? + ? = 2
3 × (−1) = −3, 3 + (−1) = 2 ✓
So we write it as 3(x + 3)(x − 1).
3(x + 3)(x − 1) = 0
A product is 0 only when one of its parts is 0.
a × b = 0 → a = 0 or b = 0
So x = −3 or x = 1.
x = −3 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) + 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 = −3, f″ is −12: negative.
A cap.
So a maximum.
f″(−3) = 6(−3) + 6 = −18 + 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″(−3) < 0 → min ✗
At x = 1, f″ is 12: positive.
A cup.
So a minimum.
f″(1) = 6(1) + 6 = 6 + 6 = 12 > 0 cup min
Its y: put −3 into f.
We get 23.
f(−3) = (−3)3 + 3(−3)2 − 9(−3) − 4 = −27 + 27 + 27 − 4 = 23
Its y: put 1 into f.
We get −9.
f(1) = 13 + 3(1)2 − 9(1) − 4 = 1 + 3 − 9 − 4 = −9
So: a maximum at (−3, 23), and a minimum at (1, −9).
max (−3, 23), min (1, −9)
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 (−3, 23), min (1, −9)
