# 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

### 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

### 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).

Solving Polynomial Inequalities

The coefficient of the highest order 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 (↗).