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

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 Link

tardymanchester tardymanchester · Answer 2 · 2019-01-21T05:39:33+01:00

Answer:

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

Step-by-step explanation:

Given : Equation [tex]y=\frac{(5-x)}{(x^2-6x+5)}[/tex]

To find :The point of discontinuity

Solution :

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

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

[tex]x^2-6x+5=0[/tex]

Solving equation by middle term split

[tex]x^2-5x-x+5=0[/tex]

[tex]x(x-5)-1(x-5)=0[/tex]

[tex](x-1)(x-5)=0[/tex]

[tex](x-1)=0,(x-5)=0[/tex]

[tex]x=1,5[/tex]

Therefore, the points of discontinuity is 1,5

but if we put in equation

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