We have $f:(0,oo)->RR;f(x)=(x+a)/(x+b);a,binRR$ .How you verify if $f$ is a bijective function?

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Answer 1 · 2017-04-29T12:39:41

f(x1) = f(x2)

$f(x_1)$ = $f(x_2)$
$(x_1+a)/(x_1+b)$ = $(x_2+a)/(x_2+b)$

Now, just do cross multiplication and simplify to obtain

$x_1=x_2$

For surjectivity,

let, y = $(x+a)/(x+b)$

Then, $yx+yb=x+a$
$\implies yb=x(y-1)+a$
$\implies(yb-a)/(y-1)+a$

Therefore, y can assume any real value except 1, which proves that it is not surjective

Answer 2 · 2017-04-30T07:28:54

See the Explanation.

It is injective, as shown by Abjo D.

But it is not surjective. Consider, $y=1 in RR.$

We claim that, corresponding to this

$y=1 in RR," there is no "x in (0,oo)," such that, "f(x)=y=1.$

Suppose, to the contrary, that $f(x)=1," for some "x in (0,oo).$

$f(x)=1 rArr (x+a)/(x+b)=1 rArr x+a=x+b, or, a=b,$ a contradiction.

Enjoy Maths.!

We have f:(0,oo)->RR;f(x)=(x+a)/(x+b);a,binRRf:(0,oo)->RR;f(x)=(x+a)/(x+b);a,binRR.How you verify if ff is a bijective function?