Question

Is the function an injection or surjection or bijection? Justify your answer. f : R → R defined by f(x) = $\frac{2x + 1}{3}$

Answer 1

Given,

f(x) = 2x + 1 / 3 given by f : R → R

Checking for injection ( One one function)

$f(x_{1} ) = \frac{ 2(x_{1} ) + 1 }{3} \\ \\ f(x_{2} ) = \frac{ 2(x_{2} ) + 1 }{3}$

Now

$\: \: \\ \\ f(x_{1}) = f(x_{2}) \\ \\ \implies \: \frac{2( x_{1}) + 1}{3} = \frac{2( x_{2}) + 1}{3} \\ \\ \implies \: 2( x_{1}) + 1\: = 2( x_{2}) + 1 \\ \\ \implies \: x_{1} = x_{2}$



As each element has a unique image, The function is an injection.

Also, Checking for surjection

Let f(x) = y

2x + 1 / 3 = y

2x + 1 = 3y

2x = 3y - 1

x = 3y - 1 / 2

As For every value of y, there exists a real value of x so It is a surjective function.


As The function is both one-one and onto, it's a bijection.

Therefore, f(x) is injection, surjection, bijection.