Worked examples
The question
Question: find the absolute maximum and minimum of f(x) = x² − 4x + 1 on [−1, 3].
f(x) = x2 − 4x + 1
On a closed interval, the highest and lowest values are at an end, or where the slope is 0.
So we test those, and only those.
candidates: the two ends + where f′ = 0
1 Critical points
Use the power rule on each term.
f′(x) = 2x^(2 − 1) − 4 × 1 + 0
Now we have f′(x) = 2x − 4.
= 2x − 4
Critical points: where the slope is 0.
Solve f′(x) = 0.
f′(x) = 0
2x − 4 = 0
Take out the 2.
2(x − 2) = 0
So x = 2.
x = 2
2 Keep the ones inside
x = 2 is inside the interval.
Keep it.
x = 2 in [−1, 3] ✓
3 Test every candidate
The left end: f of −1 is 6.
f(−1) = (−1)2 − 4(−1) + 1 = 1 + 4 + 1 = 6
The critical point: f of 2 is −3.
f(2) = 22 − 4(2) + 1 = 4 − 8 + 1 = −3
The right end: f of 3 is −2.
f(3) = 32 − 4(3) + 1 = 9 − 12 + 1 = −2
4 Biggest and smallest
Careful!
The maximum is at an end, where the slope is not 0.
That is why we test the ends.
Careful: the ends count too
The biggest value is 6, at x = −1.
That is the absolute maximum.
max : 6 at x = −1
The smallest value is −3, at x = 2.
That is the absolute minimum.
min : −3 at x = 2
5 Check
Let's check with the graph, only from −1 to 3.
The highest point is at x = −1, and the lowest at x = 2.
Right!
Let's look back.
First: critical points.
Then: keep the ones inside.
Then: test every candidate.
Then: biggest and smallest.
Last: check.
▶ Watch it solved
The question
Question: find the absolute maximum and minimum of f(x) = x³ − 3x on [0, 3].
f(x) = x3 − 3x
On a closed interval, the highest and lowest values are at an end, or where the slope is 0.
So we test those, and only those.
candidates: the two ends + where f′ = 0
1 Critical points
Use the power rule on each term.
f′(x) = 3x^(3 − 1) − 3 × 1
Now we have f′(x) = 3x² − 3.
= 3x2 − 3
Critical points: where the slope is 0.
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
2 Keep the ones inside
x = −1 is outside the interval.
Throw it away.
x = −1 not in [0, 3] ✗
x = 1 is inside the interval.
Keep it.
x = 1 in [0, 3] ✓
3 Test every candidate
The left end: f of 0 is 0.
f(0) = 03 − 3(0) = 0 + 0 = 0
The critical point: f of 1 is −2.
f(1) = 13 − 3(1) = 1 − 3 = −2
The right end: f of 3 is 18.
f(3) = 33 − 3(3) = 27 − 9 = 18
4 Biggest and smallest
Careful!
The maximum is at an end, where the slope is not 0.
That is why we test the ends.
Careful: the ends count too
The biggest value is 18, at x = 3.
That is the absolute maximum.
max : 18 at x = 3
The smallest value is −2, at x = 1.
That is the absolute minimum.
min : −2 at x = 1
5 Check
Let's check one more point: x = 1.5 gives −1.12.
It is between the min and the max.
Right!
f(1.5) = −1.12: between −2 and 18
Let's look back.
First: critical points.
Then: keep the ones inside.
Then: test every candidate.
Then: biggest and smallest.
Last: check.
▶ Watch it solved
The question
Question: find the absolute maximum and minimum of f(x) = x³ − 3x on [−2, 2].
f(x) = x3 − 3x
On a closed interval, the highest and lowest values are at an end, or where the slope is 0.
So we test those, and only those.
candidates: the two ends + where f′ = 0
1 Critical points
Use the power rule on each term.
f′(x) = 3x^(3 − 1) − 3 × 1
Now we have f′(x) = 3x² − 3.
= 3x2 − 3
Critical points: where the slope is 0.
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
2 Keep the ones inside
x = −1 is inside the interval.
Keep it.
x = −1 in [−2, 2] ✓
x = 1 is inside the interval.
Keep it.
x = 1 in [−2, 2] ✓
3 Test every candidate
The left end: f of −2 is −2.
f(−2) = (−2)3 − 3(−2) = −8 + 6 = −2
The critical point: f of −1 is 2.
f(−1) = (−1)3 − 3(−1) = −1 + 3 = 2
The critical point: f of 1 is −2.
f(1) = 13 − 3(1) = 1 − 3 = −2
The right end: f of 2 is 2.
f(2) = 23 − 3(2) = 8 − 6 = 2
4 Biggest and smallest
Careful!
The minimum is at an end, where the slope is not 0.
That is why we test the ends.
Careful: the ends count too
The biggest value is 2, at x = −1 and x = 2.
That is the absolute maximum.
max : 2 at x = −1 and x = 2
The smallest value is −2, at x = −2 and x = 1.
That is the absolute minimum.
min : −2 at x = −2 and x = 1
5 Check
Let's check with the graph, only from −2 to 2.
The highest point is at x = −1 and x = 2, and the lowest at x = −2 and x = 1.
Right!
Let's look back.
First: critical points.
Then: keep the ones inside.
Then: test every candidate.
Then: biggest and smallest.
Last: check.
Answer min : −2 at x = −2 and x = 1
▶ Watch it solved
