Question

Find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? $$ f(x)=\frac{x}{x^{2}+1} $$

Short answer

The function \(f(x)=\frac{x}{x^{2}+1}\) doesn't have any points of discontinuity in the set of real numbers. Therefore, it doesn't have any removable discontinuity.

Step-by-step solution

Identify the type of function

The function given is \(f(x)=\frac{x}{x^{2}+1}\), which is a rational function.

Find the points of discontinuity

The rational function is discontinuous where the denominator is zero. So, need to find points \(x\) where the denominator \(x^{2}+1=0\). Solving this equation doesn't yield any real solution because square of a real number is always non-negative (thus \(x^{2}+1\ge 1\)) and can't be zero. Therefore, there is no real value of \(x\) for which the function is discontinuous.

Identify removable discontinuities

As there are no points of discontinuity in the function \(f(x)=\frac{x}{x^{2}+1}\), there are no removable discontinuities.