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# In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.(i) $f : R\; \to R\; defined \; by \; f(x)\; =\; 3-4x$

Note: This is the 1st  part of a  2 part question, which is split as 2 separate questions here.

Toolbox:
• A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given $f: R \to R$ defined by $f(x)=3-4x$
$x_1\,x_2$ be two elements $\in R$
Step1: Injective or One-One function:
$f(x_1)=f(x_2)$
$=>3-4x_1=3-4x_2$
$4x_1=4x_2$
$x_1=x_2$
$f: R \to R$ defined by $f(x)=3-4x$ is one one
Step 2: Surjective or On-to function:
For any element y in R then exist a real number in R such that $f(x)=y$
Let x=$\frac{3-y}{4}$
$f(\frac{3-y}{4})=3-4(\frac{3-y}{4})=y$
$f: R \to R$ defined by $f(x)=3-4x$ is onto since ,A function $f:A \to B$ is onto if for $y \in B \; then\;exist \; x \in A$ such that $f(x)=y$
Solution: Hence $f: R \to R$ defined by $f(x)=3-4x$ is bijection

edited Mar 20, 2013 by meena.p