Question

show that f:Q–>Q is defined by f(x)= 5x+4 is bijection and find f^-1​

Answer 1

Given:

f: Q -> Q defined $\[f(x) = 5x + 4\]$

To Show:

We have to show that given function is bijection . Hence find f^-1.

Solution:

To show given function is bijection ,we have to show it one-one and onto both.

Claim: f is one-one.

Let x and y be two elements of the domain (Q), we have

$\[\begin{gathered} f(x) = f(y) \hfill \\ \Rightarrow 5x + 4 = 5y + 4 \hfill \\ \Rightarrow 5x = 5y \hfill \\ \Rightarrow x = y \hfill \\ \end{gathered} \]$

Therefore, f is one-one.

Claim: f is onto.

Let y belongs to co-domain (Q), we have

$\[\begin{gathered} f(x) = y \hfill \\ \Rightarrow 5x + 4 = y \hfill \\ \Rightarrow 5x = y - 4 \hfill \\ \Rightarrow x = \frac{{y - 4}}{5} \hfill \\ \end{gathered} \]$

x belongs to domain (Q)

Therefore, f is onto.

Thus, f is bijection and hence there exist f^-1.

To Find: f^-1  

Let,

$\[\begin{gathered} {f^{ - 1}}(x) = y \hfill \\ x = f(y) \hfill \\ x = 5y + 4 \hfill \\ 5y = x - 4 \hfill \\ y = \frac{{x - 4}}{5} \hfill \\ So,{f^{ - 1}}(x) = \frac{{x - 4}}{5} \hfill \\ \end{gathered} \]$

Answer 2

SOLUTION :

GIVEN

A function f : Q–>Q is defined by f(x)= 5x+4

TO EVALUATE

1. f is bijection

$\sf{2. \: \: To \: find \: \: {f}^{ - 1} \: \: }$

EVALUATION

CHECKING FOR ONE TO ONE

$\sf{Let \: x, y \in \mathbb{Q} \: \: such \: that \: f(x) = f(y) }$

Now f(x) = f(y) gives

$\sf{5x + 4 = 5y + 4 \: }$

$\implies \sf{5x = 5y}$

$\implies \sf{x = y}$

Hence f is one to one

CHECKING FOR ONTO

$\sf{Let \: y \in \mathbb{Q} }$

$\sf{If \: possible \: there \: exists \: x \in \mathbb{Q}} \: such \: that \: f(x) = y$

Which gives

$\sf{ 5x + 4 = y\: }$

$\implies \displaystyle \sf{ x = \frac{y - 4}{5} \: }$

$\displaystyle \sf{ As \: \: y \in \mathbb{Q} \: \: \: so \: \: \frac{y - 4}{5} \:\in \mathbb{Q} }$

Since y is arbitrary

So f is onto

Hence f is bijection

$\sf{Hence \: \: {f}^{ - 1} \: \: exists}$

DETERMINATION OF INVERSE OF THE FUNCTION

$\displaystyle \sf{ let \: \: {f}^{ - 1}(x) = y \: }$

$\implies \displaystyle \sf{ x =f(y) \: }$

$\implies \displaystyle \sf{ x = 5y + 4 \: }$

$\implies \displaystyle \sf{ y = \frac{x - 4}{5} \: }$

So

$\displaystyle \sf{ {f}^{ - 1} (x)= \frac{x - 4}{5} \: }$

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