Question

State which of the following are one - one and onto functions
i. f:
RR defined by f(x) = 2x + 1.​

Answer 1

The function $f:\mathbb{R}\to\mathbb{R}$ is defined as,

$\longrightarrow f(x)=2x+1$

We've to check if it's one - one, onto or both.

A real function $f:A\to B$ is said to be one - one if and only if all elements in $A$ has distinct images in $B,$ i.e., $f(a_1)\neq f(a_2)\ \iff\ a_1\neq a_2\ \ \forall\,a_1,\ a_2\in A.$

Assume that,

$\longrightarrow f(x_1)\neq f(x_2)$

By definition,

$\longrightarrow 2x_1+1\neq2x_2+1$

Subtracting 1,

$\longrightarrow 2x_1\neq2x_2$

Dividing by 2,

$\longrightarrow x_1\neq x_2$

This implies $f(x)$ is a one - one function.

A real function $f:A\to B$ is said to be onto if and only if the range is the codomain $B$ itself, i.e., every elements in $B$ have preimages in $A.$

Here,

$\longrightarrow f(x)=2x+1$

$\longrightarrow 2x=f(x)-1$

$\longrightarrow x=\dfrac{f(x)-1}{2}$

Here $f(x)$ can accept all real number values without any restriction, hence $f(x)\in\mathbb{R},$ i.e., range of $f$ is the codomain itself.

This implies $f(x)$ is an onto function.

Therefore, $f(x)$ is both one - one and onto.

Answer 2

Step-by-step explanation:

It is both one-one and onto function.