Question

Fine the end behavior and x-value of holeEquation is to the right

Answer 1

Given the function:

$\: f\mleft(x\mright)=\frac{x+1}{\left(x-3\right)\left(x^2-1\right)}$

The end behaviour of the function f(x) describes the behaviour of the function as x approaches +∞ and -∞.

When x approaches +∞,

$\lim _{x\rightarrow\infty}f(x)=\text{ }\lim _{x\rightarrow\infty}\frac{x+1}{(x-3)(x^2-1)}=0$

Thus, as x approaches +∞, the function f(x) approaches zero.

When x approaches -∞,

$\lim _{x\rightarrow-\infty}f(x)=\text{ }\lim _{x\rightarrow-\infty}\frac{x+1}{(x-3)(x^2-1)}=0$

Thus, as x approaches -∞, the function f(x) approaches zero.

x-value of the hole:

For a rational function f(x) given as

$f(x)=\frac{p(x)}{q(x)}$

Provided that p(x) and q(x) have a common factor (x-a), the function f(x) will have a hole at x=a.

Thus, from the function f(x)

$f(x)=\frac{x+1}{(x-3)(x^2-1)}$

by expansion, we have

$f(x)=\frac{x+1}{(x-3)(x^{}-1)(x+1)}$

The expression (x-1) is a common factor of the numerator and the denominator.

Thus,

$\begin{gathered} x-1=0 \\ \Rightarrow x=1 \end{gathered}$

Hence, the x-value of the hole is 1.