Question

Let $r(x) = \frac{8x-x^{2} }{x^{4}-64x^{2}}$.

Find the hole in the graph of y = r(x).

Answer 1

Answer:

x=8

Step-by-step explanation:

Discontinuity of a Function

We can find some functions whose graphs cannot be plotted in one stroke. It can be a hole or a vertical asymptote or a jump. To find a possible hole in a rational function, we must set both numerator and denominator to 0 independently. If a common point is found, it's a candidate for a hole if the function could eventually be redefined as continuous.

Let's find the zeros of the numerator

$8x-x^2=0$

Factoring

$x(8-x)=0$

We find two solutions: x=0, x=8

Let's find the zeros of the denominator

$x^4-64x^2=0$

Factoring

$x^2(x-8)(x+8)=0$

We find three roots: x=0, x=8, x=-8

There are two common points where the function can have holes, those are

$x=0,\ x=8$

We are not sure if those values are holes or not until we find the limits

$\displaystyle \lim\limits_{x \rightarrow 8}\frac{x(8-x)}{x^2(x-8)(x+8)}$

Simplifying

$\displaystyle =\lim\limits_{x \rightarrow 8}-\frac{1}{x(x+8)}$

$\displaystyle =-\frac{1}{128}$

Since the limit exists, the function can be redefined to cover up the hole. Now let's find the limit in x=0

$\displaystyle \lim\limits_{x \rightarrow 0}\frac{x(8-x)}{x^2(x-8)(x+8)}$

Simplifying

$\displaystyle =\lim\limits_{x \rightarrow 0}-\frac{1}{x(x+8)}$

$\displaystyle =-\frac{1}{0}=-\infty$

The limit does not exist and goes to infinity, it's not a hole, thus the only hole occurs when x=8