Question

In Exercises \(1-10,\) find the points of continuity and the points of discontinuity of the function. Identify each type of discontinuity. $$y=\frac{1}{x^{2}+1}$$

Short answer

The function \(y=\frac{1}{x^{2}+1}\) is continuous at all points for real values of \(x\), and there are no points of discontinuity.

Step-by-step solution

Identify function domain

The first step in this exercise is to identify the domain of the function \(y=\frac{1}{x^{2}+1}\). A function is continuous at a particular point if the function is defined for that point. So, we first look for values that make the denominator equal to zero, as dividing by zero is undefined. For the given function, \(x^{2}+1=0\) is the condition that would make the denominator zero. However, there is no real value of \(x\) that can make this equation true since a square of any real number is always non-negative. So the function is defined for all real values of \(x\)

Identify points of continuity or discontinuity

Given that the function is defined for all real numbers, we can say that the function is continuous at all points in its domain. As there are no points where \(x^{2}+1=0\), there are no points of discontinuity either.

Identify type of discontinuity

As we already mentioned in step 2, there are no points of discontinuity for the function. Hence it's unnecessary to identify any type of discontinuity.