If f: R rightarrow R is defined by f(x)=2x+3, | vedclass

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(A) Let $x_{1}, x_{2} \in R$. For injectivity,let $f(x_{1}) = f(x_{2})$. $2x_{1} + 3 = 2x_{2} + 3$ $2x_{1} = 2x_{2}$ $x_{1} = x_{2}$. Hence,$f$ is inj

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is given by $\frac{x-3}{2}$

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(A) Let $x_{1}, x_{2} \in R$. For injectivity,let $f(x_{1}) = f(x_{2})$. $2x_{1} + 3 = 2x_{2} + 3$ $2x_{1} = 2x_{2}$ $x_{1} = x_{2}$. Hence,$f$ is injective. For surjectivity,let $y \in 	ext{codomain } R$. Let $y = f(x) = 2x + 3$. $y - 3 = 2x$ $x = \frac{y-3}{2}$. Since for every $y \in R$,there exists an $x = \frac{y-3}{2} \in R$,$f$ is surjective. Since $f$ is both injective and surjective,$f^{-1}$ exists. Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x-3}{2}$.

If $f: R \rightarrow R$ is defined by $f(x)=2x+3$,then $f^{-1}(x)$

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