can a removable discontinuity be a local maximum

Mac Grediser

2 min read

·

27-11-2024

1

0

Share

Can a Removable Discontinuity Be a Local Maximum?

The question of whether a removable discontinuity can be a local maximum is a subtle one, requiring a careful examination of the definitions involved. Intuitively, it seems unlikely: a local maximum implies a point where the function's value is greater than its neighbors, but a removable discontinuity suggests a "hole" in the graph, potentially leaving space for higher values nearby. Let's delve into the details.

Understanding the Definitions:

• Removable Discontinuity: A removable discontinuity occurs at a point where the function is undefined or has a different value than its limit. This "hole" can be "filled" by redefining the function at that single point. The limit exists at the point of discontinuity, but it's not equal to the function's value (or the function is undefined at that point).

• Local Maximum: A function f(x) has a local maximum at x = c if f(c) ≥ f(x) for all x in some open interval containing c. In simpler terms, it's a point where the function value is higher than the values immediately surrounding it.

The Crucial Point: The Function's Definition

The key lies in how we define the function at the point of discontinuity. Let's consider a hypothetical function:

f(x) = { x^2  if x ≠ 2
       { 1   if x = 2

At x = 2, we have a removable discontinuity. The limit as x approaches 2 is 4 (lim_x→2 f(x) = 4), but f(2) = 1. The graph would show a "hole" at (2, 4) and a point at (2, 1).

In this case, the point (2, 1) is not a local maximum. The function values near x = 2 are greater than 1 (they approach 4). Therefore, there exists an interval around x = 2 where the function values are larger than f(2).

However, consider this modified function:

g(x) = { x^2  if x ≠ 2
       { 5   if x = 2

Here, we've redefined the function at x = 2. Now, g(2) = 5. In a small interval around x = 2, the value of g(x) (approaching 4) is less than g(2) = 5. Therefore, (2, 5) could be considered a local maximum. Crucially, we've redefined the function to create a local maximum at the point of the former removable discontinuity.

Conclusion:

A removable discontinuity itself cannot be a local maximum. However, by appropriately redefining the function at the point of discontinuity, we can create a local maximum at that point. The existence of a local maximum depends entirely on the function's definition, not solely on the presence of the removable discontinuity. The discontinuity simply highlights a potential point where a local maximum could be constructed by altering the function's value at that specific point. The limit of the function at the point of discontinuity must be less than the redefined value for a local maximum to exist.