Question

In Exercises \(75-86\), use a computer algebra system to analyze the graph of the function. Label any extrema and/or asymptotes that exist. $$ f(x)=\frac{1}{x^{2}-x-2} $$

Short answer

The function \(f(x)=\frac{1}{x^{2}-x-2}\) has two vertical asymptotes at \(x_{1} = 2\) and \(x_{2} = -1\) and a horizontal asymptote at \(y = 0\). There is a local minimum at \(x=\frac{1}{2}\).

Step-by-step solution

Analyze the denominator

Set the denominator equal to zero and solve for x. This will give the vertical asymptotes. So, \(x^{2}-x-2 = 0\). Factorizing the quadratic equation gives: \(x^{2}-2x+x-2 = 0\). Grouping the terms, then factoring give \((x-2)(x+1) = 0\). Therefore, \(x_{1} = 2\) and \(x_{2} = -1\) are the vertical asymptotes.

Calculating derivative

Find the derivative of the function \(f(x)\) using the quotient rule, which is necessary to find the extrema. This involves differentiating the numerator and the denominator separately, then applying the quotient rule as follows \(f'(x) = \frac{(x^{2}-x-2)'(1)-(1)'(x^{2}-x-2)}{(x^{2}-x-2)^{2}} = \frac{(2x-1)(1)-0}{(x^{2}-x-2)^{2}}= \frac{2x-1}{(x^{2}-x-2)^{2}}\). The extrema can be found by setting the derivative equal to zero and solve for x. This is \(\frac{2x-1}{(x^{2}-x-2)^{2}} = 0\), which gives \(x=\frac{1}{2}\). To determine whether this is a maximum or a minimum, the second derivative test can be used, but in this case since the function goes to \(∞\) when \(x\) goes to either extremum, it is clear that \(x=\frac{1}{2}\) is a local minimum.

Determining the horizontal asymptote

The horizontal asymptote of the function can be determined by looking at the degrees of the numerator and denominator. If the degree of the denominator is higher than the degree of the numerator, there is a horizontal asymptote at \(y = 0\). This is the case here.