Worked examples
The question
Question: find where f(x) = x² − 6x + 5 is increasing and where it is decreasing.
f(x) = x2 − 6x + 5
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) − 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
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 cut the line into 2 parts.
Test one number in each part: 2, 4.
f′(2) = −2, f′(4) = 2
f′(2) is −2: negative.
So f goes down there.
f′(4) is 2: positive.
So f goes up there.
0
4 Read the table
f is increasing, going up, where f′ is positive: x > 3.
up ↗ : x > 3
f is decreasing, going down, where f′ is negative: x < 3.
down ↘ : x < 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: read the table.
Last: check.
▶ Watch it solved
The question
Question: find the maximum and minimum points of f(x) = x³ − 3x.
f(x) = x3 − 3x
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 × 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
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 −1 and 1 cut the line into 3 parts.
Test one number in each part: −2, 0, 2.
f′(−2) = 9, f′(0) = −3, f′(2) = 9
f′(−2) is 9: positive.
So f goes up there.
f′(0) is −3: negative.
So f goes down there.
f′(2) is 9: positive.
So f goes up there.
0
0
4 Max and min
At x = −1, 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 −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.
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 (−1, 2), min (1, −2)
▶ Watch it solved
The question
Question: A rectangle has a perimeter of 40 m.
Find the largest area.
1 Name the unknown
Call the width x.
x = the width
The perimeter goes round all 4 sides: width and length, twice.
So width plus length is half of 40: 20.
width + length = 40 ÷ 2 = 20
So the length is 20 − x.
20 − x = the length
2 Write the area
Area is one side times the other.
A(x) = x(20 − x)
The x in front of the bracket multiplies every part inside.
Now we have A(x) = 20x − x².
= 20x − x2
3 Solve f′(x) = 0
The biggest area is at the top of the curve, where the slope is 0.
The power rule gives A′(x) = 20 − 2x.
A′(x) = 20 − 2x
Set it equal to 0.
20 − 2x = 0
Add 2x to both sides, so they stay equal.
2x = 20
To remove times 2, we divide by 2.
x = 202
20 divided by 2 is 10.
x = 10
4 Is it the biggest?
Before 10 the slope is positive: going up.
After it, negative: going down.
Up, then down: a maximum.
A′(9) = 2 > 0 ↗, A′(11) = −2 < 0 ↘
Careful!
x is only one side.
The question asks for the area.
Careful: x = 10 is not the answer yet
5 Answer the question
The other side is 10.
length = 20 − 10 = 10
So the largest area is 100 m².
A(10) = 10 × 10 = 100 m²
6 Check
Let's check with x a bit smaller: 9 gives 99.
x = 9: 9 × 11 = 99
And x = 11 gives 99.
Both are smaller than 100.
Right!
x = 11: 11 × 9 = 99
Let's look back.
First: name the unknown.
Then: write the area.
Then: solve f′(x) = 0.
Then: is it the biggest?.
Then: answer the question.
Last: check.
Answer A(10) = 10 × 10 = 100 m²
▶ Watch it solved
