Question

If Q is the set of rational number and f : Q → Q is defined by f(x) = 3x + 8 then prove that f is bijective. Also find f −1 .

Answer 1

$\large\underline{\sf{Solution-}}$

We know, that function is bijective if it is one - one as well as onto.

So, to show that f(x) = 3x + 8 is bijective, we have to show that f is one - one as well as onto.

One - One

$\rm :\longmapsto\:Let \: x,y \: \in \: Q \: such \: that$

$\rm :\longmapsto\:f(x) = f(y)$

$\rm :\longmapsto\:3x + 8 = 3y + 8$

$\rm :\longmapsto\:3x = 3y$

$\rm :\longmapsto\:x = y$

$\bf\implies \:f \: is \: one - one$

Onto -

Let if possible there exist an element y belongs to Q such that f(x) = y.

$\rm :\longmapsto\:3x + 8 = y$

$\rm :\longmapsto\:3x= y - 8$

$\rm :\longmapsto\:x = \dfrac{y - 8}{3}$

$\rm :\implies\:As \: y \: \in \: Q \: so \: x \: \in \: Q$

$\rm :\implies\:f \: is \: onto$

Since,

$\rm :\longmapsto\:f \: is \: one - one \: as \: well \: as \: onto$

$\rm :\implies\:f \: is \: bijective$

$\rm :\implies\: {f}^{ - 1} \: exist$

and

$\rm :\longmapsto\: {f}^{ - 1}(x) = \dfrac{x - 8}{3}$