Recently, a mathematics teacher of mine proposed a question: if there is one, what is the  the absolute minimum of the function f(x) = 1/x on the interval [0, infinity].

The answer is simple; there is none.

When considering the definition of a global minimum, one sees that a global minimum is a point of a function where f(x) is lower than any other point on the function. Stated mathematically:

f(x*) ≤ f(x)

Where x* denotes a critical number where a global minimum occurs.

In the case of f(x) = 1/x on the interval [0, infinity], as x gets closer to infinity, x continually decreases inching closer to zero. It never gets to zero but it never stops declining.

Another way to approach this would be to use derivatives. The derivative of a function is basically a function of the slope. For a function continuous on the interval [0,infinity] as in our case to have an absolute minimum, we must have the slope of that function be zero at some point in the interval. This is because for the point to be at the minimum, there must be a place where it “turns back up” so that it stops declining – the slope goes from negative to positive briefly becoming zero. That turning point is where the slope is zero. Now, using calculus, we can find that the function of 1/x on the interval [0, infinity] has no turning point and therefore cannot have a absolute minimum.

f(x) =  1 / x

f ‘(x) = -1 / x^2

0 = -1 / x^2

0 ≠ -1

Therefore, there are no “turning” points for the function 1/x on the interval [0, infinity]. With this lack of turning points, we can assume that there are no absolute extrema.