Answer:
1,5 are the points of discontinuity and 5 is the removable discontinuity.
Step-by-step explanation:
Given : Equation ![y=\frac{(5-x)}{(x^2-6x+5)}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B%285-x%29%7D%7B%28x%5E2-6x%2B5%29%7D)
To find :The point of discontinuity
Solution :
Step 1 : Write the equation ![y=\frac{(5-x)}{(x^2-6x+5)}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B%285-x%29%7D%7B%28x%5E2-6x%2B5%29%7D)
Step 2: To find the point of discontinuity we put denominator =0
![x^2-6x+5=0](https://tex.z-dn.net/?f=x%5E2-6x%2B5%3D0)
Solving equation by middle term split
![x^2-5x-x+5=0](https://tex.z-dn.net/?f=x%5E2-5x-x%2B5%3D0)
![x(x-5)-1(x-5)=0](https://tex.z-dn.net/?f=x%28x-5%29-1%28x-5%29%3D0)
![(x-1)(x-5)=0](https://tex.z-dn.net/?f=%28x-1%29%28x-5%29%3D0)
![(x-1)=0,(x-5)=0](https://tex.z-dn.net/?f=%28x-1%29%3D0%2C%28x-5%29%3D0)
![x=1,5](https://tex.z-dn.net/?f=x%3D1%2C5)
Therefore, the points of discontinuity is 1,5
but if we put in equation
[/tex]
Since (x-5) factor cancel out in the numerator and denominator therefore, it is a removable discontinuity.
And (x-1) is a infinity discontinuity.
Only (x-5) is removable discontinuity.