Question

Show that the function f:R-{1}->R - {2} defined by f(x)=(2x-3)/(x-1) is a bijection and find inverse function Show that the function f:R-{1}->R - {2} defined by f(x)=(2x-3)/(x-1) is a bijection and find inverse function @Mathematics

Answer 1

solve
\[y=\frac{2x-3}{x-1}\] for x, or solve
\[x = \frac{2y-3}{y-1}\] for y

Answer 2

Yeah I know how to find the inverse just confused about f:R-{1}->R - {2}

Answer 3

basically if you can solve you have proved that it is a bijection, so long as
\[x\neq 1\] and
\[y\neq 2\]

Answer 4

first of all the domain of
\[f(x)=\frac{2x-3}{x-1}\] must exclude 1 because you cannot divide by 0

Answer 5

and the range must exclude 2 because a fraction is only two if the numerator is twice the denominator, which is never true in this case.

Answer 6

i suppose if you like you can start with
\[f(a)=f(b)\] and use algebra to show that
\[a=b\]
this will show that the function is one to one
but this seems unnecessary if you have already found the inverse.

Answer 7

Oh okay! Thanks a lot for your help! Very detailed

Answer 8

in any case your inverse is
\[f^{-1}(x)=\frac{x-3}{x-2}\] which clearly excludes 2 right?

Answer 9

yw

Answer 10

Yeah but can you quickly explain what it means by this? f:R-{1}->R - {2}

Answer 11

R is the set of real numbers
R-{1} is the set of real numbers excluding 1
R-{2} is the set of real numbers excluding 2

Answer 12

it is saying for any number we plug in (not including 1 since 1 is not in the domain) we will get some real number that's not 2

Answer 13

Thank you! Makes a lot more sense