Question

For each function, determine the long run behavior:

x^3+1 divided by x^2+1


x^2+1 divided by x^2+2

Answer 1

The long-run behavior of the function
`f(x)=(x^3+1)/(x^2+1)` as
`x\rightarrow -\infty`,
`f(x)\rightarrow -\infty` and as
`x\rightarrow +\infty`,
`f(x)\rightarrow +\infty`.

The long-run behavior of the function
`g(x)=(x^2+1)/(x^2+2)` as

`x\rightarrow \pm \infty`,
`g(x)\rightarrow 1`.

Explanation:

The long-run behavior of rational functions refers to what happens to the graph of the function as the x-values get big or small.

Suppose that
`f(x)=(P(x))/(Q(x))` is a rational function. Let
`ax^n` be the leading term of P(x) and
`bx^m` be the leading term of Q(x). Then the long run behavior of f(x) is the same as the long run behavior of the power function

`(ax^n)/(bx^m)=(a)/(b)x^(n-m)`

We have the following functions:

(a)
`f(x)=(x^3+1)/(x^2+1)`

The leading term of the numerator is
`x^3` and the leading term of the denominator is
`x^2`. So, the long run behavior of f(x) is given by

`f(x)=(x^3+1)/(x^2+1)=(x^3)/(x^2)=x`

This tells us that when the inputs get extreme (i.e.,
`x\rightarrow \pm \infty`) the outputs will be

As
`x\rightarrow -\infty`,
`f(x)\rightarrow -\infty` and as
`x\rightarrow +\infty`,
`f(x)\rightarrow +\infty`

(b)
`g(x)=(x^2+1)/(x^2+2)`

We note that the leading term of the numerator is
`x^2` and the leading term of the denominator is
`x^2`. So, the long run behavior of g(x) is given by

`g(x)=(x^2+1)/(x^2+2)=(x^2)/(x^2)=1`

This tells us that when the inputs get extreme (i.e.,
`x\rightarrow \pm \infty`) the outputs will be about 1. This means that the graph of g(x) gets closer and closer to the line y = 1 as
`x\rightarrow \pm \infty`.