Question

how can I determine if the following function is injective, sobrejective, or biyective.

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

Can you show the work please

Answer 1

injective is easy to tell
does it pass the horizontal line test?

Answer 2

I can figure out how the graph is mentally. however, I need to do it mathematically

Answer 3

oh i see

Answer 4

for example in one exam, how can I prove that that function is one of the above menciones

Answer 5

so you want to just do it purely my definition

Answer 6

yeah. havent found exercises like that

Answer 7

we know it is 1 to 1 so we cannot provide a counterexample
so proving it by definition...

Answer 8

let see what raheen has to tell us

Answer 9

we know it isn't surjective

Answer 10

so its not bijective

Answer 11

for the injective test just plug an arbitrary x (from the domain) for example x= 3 ===> y = 2/1 = 2 . now let y =2 and solve for x ( if you get the same x i.e x=3, ===> 2=(x-1)(x-2) ==> x= 3 ,then it's injective

Answer 12

counterexample for surjective
find inverse
so x=(2y-1)/(y-1)
y cannot be 1

Answer 13

or i mean y is never 1

Answer 14

not every real number was hit

Answer 15

i think i got it

Answer 16

from the injective test and the surjective test we know if the function is bijective or not.

Answer 17

injective:
Let a, b be real numbers such that f(a)=f(b) => (a-1)/(a-2)=(b-1)/(b-2)
this is only true when a=b
so this implies f(x)=(x-1)/(x-2) is injective

Answer 18

is is special case then?

Answer 19

example of a non 1-1 function is f(x)=x^2
since:
let a,b be real numbers such that f(a)=f(b)=>
a^2=b^2
this is true when a=b or when a=-b
so since we have a being two different numbers
then f(x)=x^2 is not 1 to 1