Question

In Exercises \(21-24\), determine whether the function has a vertical asymptote or a removable discontinuity at \(x=-1 .\) Graph the function using a graphing utility to confirm your answer. $$ f(x)=\frac{x^{2}-1}{x+1} $$

Short answer

The function \(f(x)=\frac{x^{2}-1}{x+1}\) has a removable discontinuity at \(x = -1\), and a hole at \((-1, -2)\). It does not have a vertical asymptote at \(x = -1\).

Step-by-step solution

Simplifying the function

Let's begin by simplifying the function. \(f(x) = \frac{x^{2}-1}{x+1}\) can be rewritten as \(f(x) = \frac{(x-1)(x+1)}{x+1}\). This simplifies further to \(f(x) = x - 1\) since the \(x + 1\) in the numerator and denominator cancel each other out. This would not, however, apply at \(x = -1\) because in the original function that would make the denominator zero.

Checking for discontinuity

Now we have to check what happens at \(x = -1\). If we substitute this value into the simplified equation we get \(f(-1) = -2\), but if we substitute into the original function the denominator becomes zero. This indicates a removable discontinuity at \(x = -1\) and also a 'hole' at the point \((-1, -2)\) where the function is undefined.

Graphing the function

Finally, by using a graphing utility or plotting the function on graph paper, you can verify the simplified function \(f(x) = x - 1\) has a hole at \((-1, -2)\), therefore confirming the answer.