Question

What are the points of discontinuity? Are they all removable? Please show your work. y= (5-x)/x^2-6x+5

Answer 1

We have the following equation:
 y = (5-x) / x ^ 2-6x + 5
 Rewriting we have:
 y = (5-x) / (x-5) * (x-1)
 The discontinuity points are:
 x = 5
 x = 1
 x = 5 is removable. For this, we rewrite the function again:
 y = (-1 * (5-x)) / (- 1 * (x-5) * (x-1))
 y = ((x-5)) / ((x-5) * (1-x))
 We cancel similar terms:
 y = 1 / (1-x)
 Answer:
 the points of discontinuity are:
 x = 5
 x = 1
 x = 5 is removable

Answer 2

Answer:

1,5 are the points of discontinuity and 5 is the removable discontinuity.

Step-by-step explanation:

Given :    Equation   $y=\frac{(5-x)}{(x^2-6x+5)}$

To find :The point of discontinuity  

Solution :

Step 1 : Write the equation  $y=\frac{(5-x)}{(x^2-6x+5)}$

Step 2: To find the point of discontinuity we put denominator =0

$x^2-6x+5=0$

Solving equation by middle term split

$x^2-5x-x+5=0$

$x(x-5)-1(x-5)=0$

$(x-1)(x-5)=0$

$(x-1)=0,(x-5)=0$

$x=1,5$

Therefore, the points of discontinuity is 1,5

but if we put in equation

$y=\frac{(5-x)}{(x^2-6x+5)}= [tex]y=\frac{(5-x)}{(x-1)(x-5)}$[/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.