5 min readβ’january 29, 2023

The Candidates Test is a useful tool for determining the absolute (global) extrema of a function. Absolute extrema are the maximum and minimum values of a function over its entire domain. These values are important to understand because they provide a comprehensive picture of how the function behaves over its entire range.

The Candidates Test involves finding the critical points of a function and then testing those critical points to determine if they correspond to absolute extrema. The critical points of a function are the points where the first derivative of the function is equal to zero or the points where the first derivative is undefined.

The following steps outline the process for using the Candidates Test to determine absolute extrema:

- Find the critical points of the function: To do this, take the first derivative of the function and set it equal to zero. This will give you the critical points.
- Evaluate the function at the critical points: This will give you the y-values of the critical points.
- Evaluate the function at the endpoints of the interval: This will give you the y-values of the endpoints.
- Compare the y-values of the critical points and endpoints to determine the absolute extrema: If the y-value of a critical point is larger or smaller than the y-values of the endpoints, then it corresponds to an absolute maximum or minimum.

The Candidates Test is particularly useful when a function cannot be expressed in closed form, or when it is difficult to find the critical points. The test involves evaluating the function at specific points, called candidates, and then determining the maximum or minimum values of the function at those points. These maximum or minimum values can then be compared to the global maximum or minimum values to determine the absolute extrema.

It is important to note that the candidates used in the Candidates Test should be chosen carefully. The candidates should include all possible points where the function could reach its maximum or minimum values, such as endpoints of the interval and any points of inflection. By including these points in the evaluation, we can be confident that we have found the absolute extrema of the function. Additionally, it is possible to use the First or Second Derivative Tests in conjunction with the Candidates Test to confirm the results and provide a more complete picture of the function.

Example 1:
Consider the function f(x) = x^3. The first derivative is f'(x) = 3x^2. Setting f'(x) = 0, we get x = 0 as the critical point. The y-value of this critical point is f(0) = 0^3 = 0. The interval for this function is (-infinity, infinity), and the endpoints are (-infinity) and (infinity). We can see that the y-value of the critical point is not larger or smaller than the y-values of the endpoints, so it does not correspond to an absolute maximum or minimum.

Example 2:
Consider the function f(x) = -x^2 + 4. The first derivative is f'(x) = -2x. Setting f'(x) = 0, we get x = 0 as the critical point. The y-value of this critical point is f(0) = -0^2 + 4 = 4. The interval for this function is (-infinity, infinity), and the endpoints are (-infinity) and (infinity). We can see that the y-value of the critical point is larger than the y-values of the endpoints, so it corresponds to an absolute maximum.

Example 3:
Consider the function f(x) = x^2. The first derivative is f'(x) = 2x. Setting f'(x) = 0, we get x = 0 as the critical point. The y-value of this critical point is f(0) = 0^2 = 0. The interval for this function is (-infinity, infinity), and the endpoints are (-infinity) and (infinity). We can see that the y-value of the critical point is not larger or smaller than the y-values of the endpoints, so it does not correspond to an absolute maximum or minimum.

Example 4:
Consider the function f(x) = x^3 - 3x^2 + x + 2. To find the absolute extrema of this function, we first find the critical points by setting the first derivative equal to zero and solving for x:
f'(x) = 3x^2 - 6x + 1 = 0
x = 1, x = 2

Next, we evaluate the function at the critical points and at any points where the function is defined and near the critical points.
f(1) = 0, f(2) = 2

Since f(1) is the minimum value and f(2) is the maximum value, we conclude that f(1) is the absolute minimum and f(2) is the absolute maximum.

Example 5:
Consider the function f(x) = x^4 - 4x^3 + 2x^2 + 6x. To find the absolute extrema of this function, we first find the critical points by setting the first derivative equal to zero and solving for x:
f'(x) = 4x^3 - 12x^2 + 4x + 6 = 0
x = -1, x = 1, x = 3/2

Next, we evaluate the function at the critical points and at any points where the function is defined and near the critical points.
f(-1) = -2, f(1) = 4, f(3/2) = 9/16

Since f(1) is the maximum value and f(-1) and f(3/2) are both lower, we conclude that f(1) is the absolute maximum.

Example 6:
Consider the function f(x) = x^2 + 2cos(x). To find the absolute extrema of this function, we first find the critical points by setting the first derivative equal to zero and solving for x:
f'(x) = 2x - 2sin(x) = 0

Next, we evaluate the function at the critical points and at any points where the function is defined and near the critical points.
f(0) = 2, f(pi/2) = 2

Since f(0) and f(pi/2) are both the maximum value and there are no other points that are higher, we conclude that f(0) and f(pi/2) are the absolute maximum values.

In conclusion, the Candidates Test is a useful tool for determining the absolute extrema of a function. It involves finding the critical points, evaluating the function at those points and at the endpoints, and comparing the y-values to determine the absolute extrema.

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