Question

Question #bbfbd

Answer 1

Domain: `x in RR "\" {1}`

Range: `f(x) in RR "\" {0}`

Explanation:

I'm assuming that your function looks like this

`f(x) = 2/(x-1)`

The domain of the function includes any value of `x` for which `f(x)` is defined. This implies that the domain of the function will not include values of `x` for which the denominator is equal to zero.

In your case, you have

`x-1 !=0 implies x !=1`

Therefore, you can say that the domain of the function will include any value of `x in RR` with the exception of `x=1`, since that value would cause

`f(1) = 2/(1-1) = 2/0 ->` undefined

The domain will thus be `x in RR "\" {1}`.

The range of the function includes any value of `f(x)` that can be produced by plugging in the accepted values of `x`. Notice that in your case, you have no way of getting

`f(x) = 0`

That is the case because a fraction can only be equal to zero if its numerator is equal to zero.

Here the numerator of the fraction is equal to `2` regardless of the value of `x`. The range of the function will thus be `f(x) in RR "\" {0}`.

graph{2/(x-1) [-10, 10, -5, 5]}

You can use the exact same approach for the function

`f(x) = 2/x - 1`

This time, the value of `x` excluded from the domain will not be `x=1`, it will be `x=0`, since

`f(0) = 2/0 - 1 ->` undefined

The range of the function will include `f(x) = 0`, but it will no longer include `f(x) = -1`

graph{2/x - 1 [-10, 10, -5, 5]}