# Local Maximum, Local Minimum

How to find the local maximum and minimum of a function: definition, 2 examples, and their solutions.

## f'(x): 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): Minus

If f'(x) is minus,

then the graph of y = f(x)

goes downward: ↘.

## Local Maximum

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.

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

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.

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.

## Examplef(x) = x^{3} - 3x^{2} - 9x + 7

To find the zeros of y = f'(x),

find f'(x).

f(x) = x^{3} - 3x^{2} - 9x + 7

Then f'(x) = 3x^{2} - 3⋅2x^{1} - 9.

Derivative of a Polynomial

Factor f'(x) = 3x^{2} - 3⋅2x^{1} - 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.

To graph y = f(x),

use the table in the solution.

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 (↗).

## Examplef(x) = x^{4} - 4x^{3} + 10

To find the zeros of y = f'(x),

find f'(x).

f(x) = x^{4} - 4x^{3} + 10

Then f'(x) = 4x^{3} - 4⋅3x^{2}.

Factor f'(x) = 4x^{3} - 4⋅3x^{2}.

Then f'(x) = 4x^{2}(x - 3).

Find the zeros of 4x^{2}(x - 3) = 0.

Then x = 0, 3.

f'(x) = 4x^{2}(x - 3)

So y = f'(x) is y = 4x^{2}(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 x^{2} 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.

To graph y = f(x),

use the table in the solution.

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 (↗).