# Global Maximum, Global Minimum

How to find the global maximum and minimum of a function: 1 example and its solution.

## Examplef(x) = x^{3} - 3x - 1, [0, 2]

To find the global maximum and minimum values,

find the endpoints of y = f(x),

find the local maximum and minimum points,

then compare these points.

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

find f'(x).

f(x) = x^{3} - 3x - 1

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

Derivative of a Polynomial

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

Then f'(x) = 3(x + 1)(x - 1).

Find the zeros of 3(x + 1)(x - 1) = 0.

Then x = ±1.

f'(x) = 3(x + 1)(x - 1)

So y = f'(x) is y = 3(x + 1)(x - 1).

And the zeros are ±1.

Draw y = f'(x).

y = f'(x) is a parabola

that is opened upward

and that passes through -1 and +1.

The given function is in an interval [0, 2]:

0 ≤ x ≤ 2.

And the zero in this interval is

x = 1.

So make a table

that starts from x = 0,

includes x = 1,

and that ends at x = 2.

See the graph of y = f'(x)

and fill the f'(x) row.

For x = 0,

f'(x) is minus.

For 0 < x < 1,

f'(x) is minus.

For x = 1,

f'(x) = 0.

For 1 < x < 2,

f'(x) is plus.

For x = 2,

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 (↗).

x = 0 and 2 are the endpoints.

At x = -1,

f(x) changes from downward (↘) to upward (↗).

So x = 1 is the local minimum.

Then find the y values of these points.

Put x = 0, 1 and 2 into the given f(x).

f(0) = 1

f(1) = -1

f(2) = 3

Write these y values into the table.

The greatest f(x) value is 3:

at x = 2.

So the global maximum is (2, 3).

The least f(x) value is -1:

at x = 1.

So the global minimum is (1, -1).

(It's the local minimum.)

The global maximum is (2, 3).

And the global minimum is (1, -1).

To graph y = f(x),

use the table in the solution.

The graph starts from (0, 1).

For 0 < x < 1,

f(x) goes downward (↘).

(1, -1) is the local minimum.

For 1 < x < 2,

f(x) goes upward (↗).

At (2, 3),

the graph ends.

So the global maximum is (2, 3).

And the global minimum is (1, -1).