Local Maximum, Local Minimum
How to find the local maximum and minimum of a function: definitions, 2 examples, and their solutions.
Derivative and Graph
f'(x) is Plus
Recall that the derivative is the slope of the graph.
Derivative: Definition
So, if f'(x) is plus,
then the graph of y = f(x) goes upward: ↗.
f'(x) is Minus
If f'(x) is minus,
then the graph of y = f(x) goes downward: ↘.
Local Maximum
Definition
The local maximum is the point
where the graph of y = f(x) changes
from upward (↗) to downward (↘).
So it's the point
where f'(x) changes from plus to minus.
So, if f(x) is differentiable near the local maximum,
you can find the local maximum
by finding the point
where f'(x) = 0
and where f'(x) changes from plus to minus.
Rough Point
Sometimes, a local maximum can be a sharp point like this.
(It's called a rough point.)
In this case,
f(x) is not differentiable at the local maximum.
So you cannot find the local maximum
by setting f'(x) = 0.
Then, check the point
where f'(x) changes from plus to minus.
Local Minimum
Definition
The local maximum is the point
where the graph of y = f(x) changes
from downward (↘) to upward (↗).
So it's the point
where f'(x) changes from minus to plus.
So, if f(x) is differentiable near the local minimum,
you can find the local minimum
by finding the point
where f'(x) = 0
and where f'(x) changes from minus to plus.
Rough Point
Sometimes, a local minimum can be a sharp point like this.
(It's also called a rough point.)
In this case,
f(x) is not differentiable at the local maximum.
So you cannot find the local minimum
by setting f'(x) = 0.
Then, check the point
where f'(x) changes from minus to plus.
Example 1
Example
Solution
To find the zeros of y = f'(x),
find f'(x).
f(x) = x3 - 3x2 - 9x + 7
Then f'(x) = 3x2 - 3⋅2x1 - 9.
Derivative of a Polynomial
Factor f'(x) = 3x2 - 3⋅2x1 - 9.
Then f'(x) = 3(x + 1)(x - 3).
Find the zeros of 3(x + 1)(x - 3) = 0.
Then x = -1, 3.
f'(x) = 3(x + 1)(x - 3)
So y = f'(x) is y = 3(x + 1)(x - 3).
And the zeros are -1 and 3.
Draw y = f'(x).
y = f'(x) is a parabola
that is opened upward
and that passes through -1 and 3.
Make a table like this.
See the graph of y = f'(x)
and fill the f'(x) row.
For x = -1 and 3,
f'(x) = 0.
For x < -1,
f'(x) is plus.
For -1 < x < 3,
f'(x) is minus.
For x > 3,
f'(x) is plus.
Fill the f(x) row.
If f'(x) is plus,
f(x) goes upward (↗).
If f'(x) is minus,
f(x) goes downward (↘).
Then see the table.
At x = -1,
f(x) changes from upward (↗) to downward (↘).
So x = -1 is the local maximum.
At x = 3,
f(x) changes from downward (↘) to upward (↗).
So x = 3 is the local minimum.
Then find the y values of these points.
Put x = -1 and 3 into the given f(x).
Then f(-1) = 12 and f(3) = -20.
Write these y values into the table.
At (-1, 12),
f(x) changes from upward (↗) to downward (↘).
So (-1, 12) is the local maximum.
At (3, -20),
f(x) changes from downward (↘) to upward (↗).
So (3, -20) is the local minimum.
So (-1, 12) is the local maximum.
And (3, -20) is the local minimum.
Graph
To graph y = f(x),
use the table above.
For x < -1,
f(x) goes upward (↗).
(-1, 12) is the local maximum.
For -1 < x < 3,
f(x) goes downward (↘).
(3, -20) is the local minimum.
For x > 3,
f(x) goes upward (↗).
Example 2
Example
Solution
To find the zeros of y = f'(x),
find f'(x).
f(x) = x4 - 4x3 + 10
Then f'(x) = 4x3 - 4⋅3x2.
Factor f'(x) = 4x3 - 4⋅3x2.
Then f'(x) = 4x2(x - 3).
Find the zeros of 4x2(x - 3) = 0.
Then x = 0, 3.
f'(x) = 4x2(x - 3)
So y = f'(x) is y = 4x2(x - 3).
Draw y = f'(x).
Polynomial Inequality
The coefficient of the highest order term is 4.
So start from the upper right.
The exponent of (x - 3) factor is 1: odd.
So y = f'(x) passes through x = 3.
The exponent of x2 factor is 2: even.
So y = f'(x) bounces off at x = 0.
Make a table like this.
See the graph of y = f'(x)
and fill the f'(x) row.
For x = 0 and 3,
f'(x) = 0.
For x < 0,
f'(x) is minus.
For 0 < x < 3,
f'(x) is also minus.
For x > 3,
f'(x) is plus.
Fill the f(x) row.
If f'(x) is minus,
f(x) goes downward (↘).
If f'(x) is plus,
f(x) goes upward (↗).
Then see the table.
At x = 0,
f(x) changes from downward (↘) to downward (↘):
there's no change.
So x = 0 is neither the local maximum nor the local minimum.
(although it's the zero of y = f'(x))
At x = 3,
f(x) changes from downward (↘) to upward (↗).
So x = 3 is the local minimum.
x = 3 is the local minimum.
Then find the y value for x = 3.
Put x = 3 into the given f(x).
Then f(3) = -17.
Write this y value into the table.
At (3, -17),
f(x) changes from downward (↘) to upward (↗).
So (3, -17) is the local minimum.
There's no local maximum.
And (3, -17) is the local minimum.
Graph
To graph y = f(x),
use the table above.
For x < 0,
f(x) goes downward (↘).
At x = 0, (0, 10),
f(x) neither goes upward nor downward. (f'(x) = 0).
For 0 < x < 3,
f(x) goes downward (↘) again.
(3, -17) is the local minimum.
For x > 3,
f(x) goes upward (↗).