# Global Maximum, Global Minimum

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

## Example

### Example

To find the global maximum and minimum values,
find the endpoints of y = f(x)
and find the local maximum and minimum points.

### Solution

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

f(x) = x3 - 3x - 1
Then f'(x) = 3x2 - 3.

Derivative of a Polynomial

Factor f'(x) = 3x2 - 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).

### Graph

To graph y = f(x),
use the table above.

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).