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An absolute minimum value of a function f is the lowest value of f(x) for all x in the domain of f. This value is attained at least once in the domain by a specific x-coordinate, known as the global minimum. In other words, it is the smallest possible value of the function over its entire domain.

An interval is a set of real numbers that consists of all numbers between two given points, called endpoints. An interval can be closed, open, or a combination of both. A closed interval includes its endpoints and is denoted as [a, b], whereas an open interval excludes its endpoints and is denoted as (a, b). A mixed interval can be either closed at one end and open at the other, such as [a, b) or (a, b].

For a function f to have an absolute minimum value at an endpoint of an interval, the following conditions must be met: 1. The function must be continuous on the interval, meaning there are no gaps, jumps, or breaks in the curve of the function. 2. The function must attain its lowest value at one of the endpoints of the interval.

Let's consider a simple example, the quadratic function \(f(x) = (x-1)^2\) on the closed interval [0, 2]. This function is continuous over the domain of real numbers and thus continuous on the interval [0, 2]. First, let's find the values of the function at the endpoints: \(f(0) = (0-1)^2 = 1\) and \(f(2) = (2-1)^2 = 1\) Now, for any \(x\) within the interval, we have: \(f(x) = (x-1)^2 \geq 0\), since the square of any real number is non-negative. Clearly, there is no value of x from the domain of the function that results in a \(f(x)\) less than 1. Therefore, we can conclude that the function f has an absolute minimum value of 1 attained at both the endpoints x = 0 and x = 2. In this example, we can see that the function f has an absolute minimum value at not just one, but both of the endpoints of the interval [0, 2]. This demonstrates that a function can indeed have an absolute minimum value at an endpoint of an interval.