Answer:
<em>x=8</em>
Step-by-step explanation:
<u>Discontinuity of a Function</u>
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](https://tex.z-dn.net/?f=8x-x%5E2%3D0)
Factoring
![x(8-x)=0](https://tex.z-dn.net/?f=x%288-x%29%3D0)
We find two solutions: x=0, x=8
Let's find the zeros of the denominator
![x^4-64x^2=0](https://tex.z-dn.net/?f=x%5E4-64x%5E2%3D0)
Factoring
![x^2(x-8)(x+8)=0](https://tex.z-dn.net/?f=x%5E2%28x-8%29%28x%2B8%29%3D0)
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](https://tex.z-dn.net/?f=x%3D0%2C%5C%20x%3D8)
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)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim%5Climits_%7Bx%20%5Crightarrow%208%7D%5Cfrac%7Bx%288-x%29%7D%7Bx%5E2%28x-8%29%28x%2B8%29%7D)
Simplifying
![\displaystyle =\lim\limits_{x \rightarrow 8}-\frac{1}{x(x+8)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D%5Clim%5Climits_%7Bx%20%5Crightarrow%208%7D-%5Cfrac%7B1%7D%7Bx%28x%2B8%29%7D)
![\displaystyle =-\frac{1}{128}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D-%5Cfrac%7B1%7D%7B128%7D)
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)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim%5Climits_%7Bx%20%5Crightarrow%200%7D%5Cfrac%7Bx%288-x%29%7D%7Bx%5E2%28x-8%29%28x%2B8%29%7D)
Simplifying
![\displaystyle =\lim\limits_{x \rightarrow 0}-\frac{1}{x(x+8)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D%5Clim%5Climits_%7Bx%20%5Crightarrow%200%7D-%5Cfrac%7B1%7D%7Bx%28x%2B8%29%7D)
![\displaystyle =-\frac{1}{0}=-\infty](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D-%5Cfrac%7B1%7D%7B0%7D%3D-%5Cinfty)
The limit does not exist and goes to infinity, it's not a hole, thus the only hole occurs when x=8