Sometimes we take vacations, sometimes we take stay-cations. Second derivatives can be used to determine if the function will be traveling somewhere extreme or if it will travel somewhere more subdued. An extreme point may be either local or global.
Global means exactly that - a maximum value of a function is a global maximum if it's the biggest y-value that function ever hits. Similarly, a global minimum is the smallest y-value a function ever hits. Any maximum or minimum that isn't global is called local.
Here's a picture of some local and global extreme points:

It's possible for a function to not have a global maximum (or minimum).
Sample Problem
The function f (x) = ex has no global maximum. Since
,
the function f keeps growing and growing and never hits a "biggest" value.

How can a function fail to have a global maximum? One way is if the function has the whole real line to play with, and keeps increasing all the way along the line, like these:

Another way is if the function has a vertical asymptote, in which case the function keeps increasing as x approaches (but never quite reaches) a particular value:

Another way is if the function has a hole. In this case the function tries to approach a maximum, but never quite reaches it:

The ways a function could fail to have a global minimum are similar—turn the previous pictures upside down.
Each function that fails to have a global max (or min) has one of two things going on with it. Either the function is defined on an open interval (such as the whole real line), or the function is discontinuous. If we toss out all such functions and only look at a continuous function on a closed interval, we're guaranteed to find both a global max and a global min on that interval. This is a restatement of the Extreme Value Theorem.
If we're looking at a continuous function on a closed interval, there are two places a global maximum (or minimum) can occur. It can occur in the middle of the interval, in which case it must be at a critical point:

Or it can occur at an endpoint of the interval:

If the global max occurs at a critical point, it must be the critical point with the largest function value:

This gives us a nice way to find the global max and min of a continuous function f on a closed interval [a,b].
- Find the critical points of f on [a, b].
- Find the function value at all critical points and at the endpoints x = a and x = b.
- The largest function value is the global max, and the smallest function value is the global min.
This can be quicker to do than the first and second derivative tests. We don't need to check the signs of derivatives or anything like that. We know the global max (and min) of f must happen either at a critical point or at an endpoint of the interval. So we find the value of f at critical points and endpoints, and take the biggest and smallest values.
Practice:
Determine if the function shown below has a global maximum and/or a global minimum. 
| |
The function has a global maximum but no global minimum. Since 
the function f keeps decreasing, but it never hits bottom. | |
Find the global max and min of the function f (x) = x2 + 2 on the interval [ -4 , 5 ].
| |
First we find the critical points. The derivative is f '(x) = 2x which is zero only when x = 0, so that's our only critical point. Next we evaluate f at the critical point and at the endpoints of the interval. 
The global max of f on [-4,5] is y = 27, which occurs at x = 5. The global min of f on [-4,5] is y = 2, which occurs at x = 0. Remember that there's a difference between the maximum value of a function (a y-value) and where that maximum occurs (an x-value). | |
On the interval [-3,3], where is the function f (x) = x3 + 12x largest? | |
This question is asking us to find the x-value at which the global maximum of f occurs. First we find the critical points. The derivative f '(x) = 3x2 + 12 is zero when x = ± 2, so these are our critical points. Next we evaluate f at the critical points and at the endpoints of the interval. 
The maximum value of f occurs at x = 3. It's possible that you might find a critical point that's not in the interval given. If this happens, ignore the offending critical point. The function f can't possibly have its maximum at x = -2 if we're only looking at the interval 0 < x < 3. | |
On the interval [0,3], where is the function f (x) = x3 + 12x smallest? | |
We found in the example above that f has critical points at x = ± 2. However, x = -2 isn't in the interval we're looking at, so we toss it out. We only keep the critical point x = 2. We evaluate the function f at the critical points in [0,3] and the endpoints of [0,3]. 
On this interval, the function f is smallest at x = 0. | |
Determine if the function has a global maximum and/or a global minimum.

Answer
This function doesn't have a global max or a global min. It keeps increasing in one direction and decreasing in the other:

Determine if the function has a global maximum and/or a global minimum.

Answer
This function does have both a global max and a global min:
Determine if the function has a global maximum and/or a global minimum.

Answer
This function has a local min and a local max, but doesn't have a global min or a global max.
Determine if the function has a global maximum and/or a global minimum.

Answer
This function has a global min, but no global max
Determine if the function has a global maximum and/or a global minimum.

Answer
This function only has one value. Since y = 2 is both the biggest and the smallest this function ever becomes, y = 2 is both the global maximum and the global minimum.
Determine if the function has a global maximum and/or a global minimum.

Answer
This function has no global min or global max.
Determine if the function has a global maximum and/or a global minimum.

Answer
This function is trying to have a global max, but never quite reaches it. It also has no global minimum, since the function approaches zero as x approaches ± ∞, but never reaches zero.
Determine if the function has a global maximum and/or a global minimum.

Answer
This function has no global max or global min.

Find the global max and global min of the function on the specified interval.
on the interval [-1,1]
Answer
Rewrite the function to make life easier:

The derivative is
f'(x) = -(x2 + 1) - 2(2x)
which is zero when x = 0.
Now evaluate f at the critical point and the endpoints of the interval:

The global max is y = 1 and the global min is y = 0.5.
Find the global max and global min of the function on the specified interval.
f (x) = x2 + 3x on the interval [-2,3]
Answer
The derivative is
f'(x) = 2x + 3
which is zero when x = -3/2.
We evaluate f at this critcial point and at the endpoints:

On the specified interval, the global max of f is y = 18 and the global min of f is y = -2.25.
Find the global max and global min of the function on the specified interval.
f (x) = xex on the interval [-2,-1]
Answer
The derivative is f'(x) = xex + ex = ex(x + 1). The only critical point occurs at x = -1, which is also an endpoint of the interval in question. It turns out that we only need to evaluate f (x) at the endpoints.

The global maximum of f on this interval is -2e-2 and the global minimum is -e-1.
Find the global max and global min of the function on the specified interval.
f (x) = sin x on the interval [π,2π]
Answer
The derivative of f is f'(x) = cos x, whose only root on the given interval occurs at
.
We evaluate f at the critical point and at the endpoints of the interval:

The global max of f on this interval is y = 0 (which occurs at two different values of x) and the global min is y = -1.
Find the global max and global min of the function on the specified interval.

Answer
The derivative is f'(x) = x2 + 3x - 10, which factors as
f'(x) = (x + 5)(x - 2).
The critical points occur at x = -5 and at x = 2.
We evaluate f at the critical points and at the endpoints of the interval:

The global max is y = 46.5 and the global min is
.
Find where the global max and global min of the function occur (on the specified interval).
f (x) = x2ex on the interval [-5,1]
Answer
The derivative is
f'(x) = 2xex + x2ex = xex(2 + x)
so there are critical points at x = -2 and at x = 0.
We evaluate f at the critical points and endpoints (it's ok to use approximate y-values here, because we're only comparing the y-values tofigure out which x-values we want):

f has its global max when x = 0 and its global min when x = 1.
Find where the global max and global min of the function occur (on the specified interval).
f (x) = 2x2 + 3x - 2 on the interval [-2,2]
Answer
The derivative is
f'(x) = 4x + 3
which is zero when
, so this is our critical point.
We evaluate f at all the important places:

The global max of f occurs when x = 2 and the global min when
.
Find where the global max and global min of the function occur (on the specified interval).
on the interval [-π,π]
Answer
The derivative is

which has a critical point whenever

This occurs when

where k is an integer. Solving,

and so

We want only the critical points in the interval [-π,π]. These would be

Now we evaluate f at lots of places:

The global max occurs at
and at
. The global min occurs at
and at
.
Find where the global max and global min of the function occur (on the specified interval).
on the interval [-1,4]
Answer
The derivative is

This is 0 when -x3 = -1, or when x = 1. This is our only critical point, since anywhere f' is undefined, f is already undefined.
Evaluate f.

The function f achieves its global max on this interval when x = 1 and its global min when x = -1.
Find where the global max and global min of the function occur (on the specified interval).
f (x) = 3(x - 2) + 9 on the interval [-5,5]
Answer
This function is a line, so it has no critical points.
Better yet, we don't even need to evaluate f at the endpoints. Since f is a line with a slope of 3, f is an increasing function. This means f must be smallest at the start of the interval and largest at the end of the interval.
The global min of f on this interval occurs at x = -5 and the global max of f on this interval occurs at x = 5.
These problems become more complicated when we start doing something called optimization, which is the same exact math but with more words added on top.