Question

Removable and Nonremovable Discontinuities In Exercises \(35-60,\) find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? \ $$ f(x)=\left\\{\begin{array}{ll}{x,} & {x \leq 1} \\ {x^{2},} & {x>1}\end{array}\right. $$

Short answer

The function is continuous for all real numbers. There are no discontinuities.

Step-by-step solution

Check for continuity at x = 1 from the left side

On the left side of \(x = 1\), the function is defined as \(f(x) = x\). So, when \(x = 1\), \(f(x) = 1\).

Check for continuity at x = 1 from the right side

On the right side of \(x = 1\), the function is defined as \(f(x) = x^2\). So, when \(x = 1\), \(f(x) = 1^2 = 1\).

Compare the function values from both sides of x = 1

As we found in steps 1 and 2, \(f(x) = 1\) from both sides (left and right) of \(x = 1\). Since the function values are the same, there is continuous at \(x = 1\). Therefore, there is no discontinuity in this function at \(x = 1\). Hence, no removable or nonremovable discontinuity exists.

Key concepts

Piecewise Continuous Functions

A piecewise continuous function is a type of function that is defined by different expressions for different intervals of the domain. These segments ensure the function exists throughout all possible values of the input in its domain. However, due to being defined in pieces, checking the function's continuity at the endpoints of these segments is crucial.

In our example, the function \( f(x) \) is defined as follows:

• For \( x \leq 1 \), \( f(x) = x \)
• For \( x > 1 \), \( f(x) = x^2 \)

Continuity in piecewise functions involves matching the function values at the boundary points between different pieces. In simple terms, you want a smooth transition from one piece to another without any jumps or holes. In the provided solution, it was verified that both sides of \( x = 1 \) give a function value of 1, thus ensuring continuity at this critical point.

Removable Discontinuities

Removable discontinuities occur when a function is not continuous at a point, but can be made continuous by redefining the function or point. This typically happens when there is a hole in the graph of the function.

In the context of piecewise functions, you might encounter an interval boundary where the left-hand and right-hand limits exist and agree, but the function might not be defined at that very point, or it might be defined differently. By adjusting the definition slightly or filling that hole, you can "remove" the discontinuity, hence the name "removable."

However, in the given exercise, a careful examination of the boundary value \( x = 1 \) shows that the function is already continuous at this point. Both segments of the piecewise function agree on the value \( f(x) = 1 \) at \( x = 1 \). Therefore, there are no removable discontinuities.

Nonremovable Discontinuities

Nonremovable discontinuities are those that cannot be simply fixed by redefining the function or its values. This type of discontinuity often manifests as a very pronounced jump or break in the graph, where the left and right limits do not match, or one or both of them do not exist.

These discontinuities could be due to:

• Vertical asymptotes – where the function shoots off to infinity
• Sudden jumps – where the two sides of a point do not meet

In our given exercise, we analyzed the piecewise function carefully and discovered that at \( x = 1 \), the left-hand and right-hand values met correctly at the same point with \( f(x) = 1 \). Since the criteria for nonremovable discontinuities are unmet, we can conclude there are none present. This clarity ensures that all parts of the function connect smoothly, confirming the absence of any abrupt jumps or vertical asymptotes, thus highlighting the continuous nature of the function throughout its domain.