find the end - behavior of the following rational functions. $g(x)=\frac{x + 1}{(2x + 5)(x - 4)}$…

Explanation:

Step1: Expand the numerator

First, expand ((2x + 5)(x - 4)) using the FOIL - method. ((2x+5)(x - 4)=2x\times x-2x\times4 + 5\times x-5\times4=2x^{2}-8x + 5x-20=2x^{2}-3x - 20). So, (g(x)=\frac{2x^{2}-3x - 20}{x + 1}).

Step2: Use long - division or the fact that for a rational function (y=\frac{f(x)}{d(x)}) where (f(x)=a_nx^n+\cdots+a_0) and (d(x)=b_mx^m+\cdots+b_0) ((n\geq m)), we can analyze the end - behavior by considering the leading terms.

The degree of the numerator (n = 2) and the degree of the denominator (m = 1). When (x\to\pm\infty), we consider the ratio of the leading terms of the numerator and the denominator. The leading term of the numerator is (2x^{2}) and the leading term of the denominator is (x). (\lim_{x\to\pm\infty}\frac{2x^{2}-3x - 20}{x + 1}=\lim_{x\to\pm\infty}\frac{2x^{2}}{x}=\lim_{x\to\pm\infty}2x=\pm\infty). We can also use long - division: (2x^{2}-3x - 20=(x + 1)(2x-5)-15), so (g(x)=2x-5-\frac{15}{x + 1}). As (x\to\pm\infty), (\lim_{x\to\pm\infty}g(x)=\lim_{x\to\pm\infty}(2x-5-\frac{15}{x + 1})). Since (\lim_{x\to\pm\infty}\frac{15}{x + 1}=0), (\lim_{x\to-\infty}g(x)=-\infty) and (\lim_{x\to\infty}g(x)=\infty).

Answer:

(\lim_{x\to-\infty}g(x)=-\infty,\lim_{x\to\infty}g(x)=\infty)