Let be an open interval. The function is:
1.) increasing on if () ,
2.) strictly increasing on if () (,
3.) decreasing on if ( ,
4.) strictly decreasing on if () .
Functions above are called monotonic functions, that is, strictly monotonic functions.
Let be the function such that is differentiable on and continuous on .
a) The function is increasing on if .
b) The function is decreasing on if .
c) The function is constant on if .
Let be an open interval, and . Then is a critical number (or stationary point) of the function if , whereby the function is defined at .
To determine intervals in which the function is either always increasing or decreasing, first we need to find critical points by solving the equation
The domain of the function is divided into open intervals with critical numbers and on each of these intervals we need to determine the sign of the first derivative.
Example 1. Determine critical numbers and monotone intervals of the function if:
The first derivative of the function is equal to:
Critical numbers we obtain by solving the equation:
Solutions of the equation above are:
that is, and are critical numbers.
Therefore, monotone intervals are:
Now we choose any point from the each interval and determine the sign of the first derivative:
We can conclude that the function is increasing on intervals and and decreasing on .
Let be an open interval, and .
Then the function has the local minimum at point if and the local maximum at point if .
The local minimum and local maximum of a function are called local extreme values or local extrema of a function.
The necessary condition for local extrema (Fermat’s theorem)
If the function has the local minimum or maximum at point and if the function has the first derivative at point , then .
Theorem. (First derivative test)
Let be differentiable function on and a critical number.
1.) If changes sign from positive to negative around number , then is a local maximum.
2.) If changes sign from negative to positive around number , then is a local minimum.
3.) If has the same sign from both sides of , then the local extremum does not exists at .
Example 2. Determine local extrema of the function:
Calculate the first derivative:
which is equal to for . On the interval the first derivative is positive, that is, in interval the given function is increasing. The first derivative is negative on the interval , that is, the function is decreasing. Therefore, the first derivative changes sign from positive to negative. The value of the function at point is equal to:
Let be the function defined on , and . Then the function has the global minimum at point if and the global maximum at point if .
Difference between local and global extrema is that the global extrema of a function is the largest or the smallest value on its entire domain, and local extrema of a function is the largest or the smallest value in a given range of a function.
How to find global extrema of the function on the closed interval ?
1.) Find all critical numbers of the function on .
2.) Evaluate the function at the endpoints and and at all critical numbers from the step 1.)
3.) The largest of obtained values is the absolute maximum and the smallest is the absolute minimum.
Example 3. Find the global minimum and global maximum of the function on the interval if
We finding the first derivative:
Critical numbers we find by solving the equation , that is . We obtain that critical numbers are and .
Now we need evaluate the function at the endpoints of the interval and at all critical numbers. Therefore, we have:
The largest value of the function is at the point , and the smallest value of the function is at the point . Therefore, the point is the absolute minimum of the function and point is the absolute maximum of the function on the interval .
The sign of the first derivative in the neighborhood of a critical number it is not always easy to determine. Therefore, there is one more criteria for determining the character of a critical number, by using the second derivative.
Concavity and points of inflection
The function , is concave up if is increasing on .
The function , is concave down if is decreasing on .
Theorem. Let be an open interval and a function which is twice differentiable on .
1.) If , then the function is concave up on .
2.) If , then the function is concave down on .
3. ) The point , is called an inflection point of the function if exists and if concavity changes the sign at from positive to negative and from negative to positive.
That is, if is it the second derivative on the intervals and of different signs, then the function at point has an inflection point, where changes from concave up to concave down or from concave down to concave up.
For instance, the function is concave up on , because , .
The function is concave down on , because , for all .
How to find open intervals where the function is concave up or concave down?
- Find the second derivative of the function .
- Potential inflection points find by solving the equation . Check if they are in the domain of the function .
- On those intervals on which the function is concave up, otherwise, concave down. On the border between concavity intervals is the inflection point.
Example 4. Find the open intervals where the function is concave up or concave down if
Firstly, we need to find .
The second derivative is not equal to for any , however for the second derivative it is not define. Therefore, for is , that is, the function is concave down and for is , that is, the function is concave up.
Theorem. (The second derivative test)
Let the function be twice differentiable on . To find the local extrema of the function we
1.) calculate and ,
2.) find critical numbers by solving the equation ,
3.) to each critical number apply the second derivative test; if , then is the local minimum, and if , then is the local maximum. If , then the character of the point we finding by using the sign of the first derivative.
Example 5. Determine extrema of the function if
Firstly, we need to find and . We have
Critical numbers we finding by solving the equation , that is
Therefore, the critical points are and . Since the second derivative of the function is defined on and the function on , the second derivative of the function at point is equal to . The second derivative of the function at point is equal to
Now we need to find the value of the function at point , that is
From the second derivative test we conclude that the given function has the local maximum at point .