is this function continuous on its domain?

Question

is f(x)= ( x^2- 3x+ 2)/ (x-1) continuous on its domain?
what is the domain and explain any apparent problems

Answer 1

No. There is a removable discontinuity @ x=1

Answer 2

f(x) = (x^2 - 3x + 2) / (x - 1)

First off, the domain of this function is (-infinity, 1) U (1, infinity)

This function is continuous on its domain, because if we take the limit as x approaches 1:

lim (x^2 - 3x + 2) / (x - 1)
x -> 1

lim (x - 1)(x - 2) / (x - 1)
x -> 1

lim (x - 2) = 3
x -> 1

The fact that we get a value means this is a removeable discontinuity. Therefore, it is continuous everywhere else except at 1.

Answer 3

Notice the point of discontinuity at x = 1.

Since your denominator is x - 1, that means that if x = 1 the denominator would be zero. Since dividing by zero is undefined (you can't do it.), there is no value at x = 1.

This is a common type of discontinuity problem. Having a function with a variable in the denominator. Generally, to find a point of discontinuity in this problem, set the denominator equal to zero (In this case, x - 1 = 0) and solve.

Answer 4

you've the right answer (i imagine. Your placed up turned right into somewhat demanding to adhere to.). both f(x) and g(x) do not exist at x=0, so as that they haven't any cost at that aspect, and obviously you won't be able to upload issues that do not exist. Continuity (as I discovered it) is merely that there aren't any discontinuities. subsequently, i does no longer call f(x)+g(x) continuous, yet i recognize some those who might want to. It really relies upon on the definition of c on your definition of continuity. If c is ANY x, even ones no longer contained in the area, then that's no longer continuous. If c is any x contained in the area, then that is. Sorry for rambling somewhat, although. : l

Answer 5

What happens when x = 1?

Answer 6

there is a hole at x =1.
the domain is:
(negative infinity, 1) (1, infinity)