How do you find the domain and range of #f(x) = (x+2) / (x-1)#?

Question

Abigail Allen · Accepted Answer

Domain and range #(-oo, 1)U(1, +oo)#

To find the domain, it has to be seen whether f(x) exists for all values of x. If it does , its domain is all Real numbers. If it does not exists for some values of x. then those numbers are excluded from the domain. In the present case f(x) would not exist for x=1, hence its domain would be all real numbers except 1. In set notation it would be written as #(-oo, 1)U(1, +oo)#.

For range, write it as #y= (x+2)/(x-1)#. Interchange x and y and solve for y, as follows:

#x=(y+2)/(y-1) rArr y= (x+2)/(x-1)# Now the domain of this function would be the range of the original function.

It turns out that this function's domain matches that of the original function as well.

Hence range of original function would also be #(-oo, 1)U(1, +oo)#

Eli Assouline · Answer

undefined To find the domain:
- The function is defined for all real numbers except where the denominator is zero.
- Set the denominator, $ x - 1 $, not equal to zero and solve for $ x $.
- $ x - 1 \neq 0 $ leads to $ x \neq 1 $.
- Therefore, the domain is all real numbers except $ x = 1 $.

To find the range:
- As $ x $ approaches positive or negative infinity, $ f(x) $ approaches 1.
- There is a vertical asymptote at $ x = 1 $.
- The range is all real numbers except $ f(x) $ cannot equal 1.