Show that the function f in A = |R - 2 3 defined as f(x) = 4x + 3 6x - 4 is one-one and onto. Hence find f^ -1 .

Let x_1, x_2 A Now f(x_ 1 ) = f(x_ 2 ) = 4x_ 1 + 3 6x_ 1 - 4 = 4x_ 2 + 3 6x_ 2 - 4 24x_ 1 x_ 2 + 18 x_ 2 - 16x_ 1 - 12 = 24 x_ 1 x_ 2 + 18 x_ 1 - 16 x_ 2 - 12 - 34 x_ 1 = - 34x_ 2 _ 1 =x_ 2 Hence f is one-one function For onto Let y = 4x + 3 6x - 4 6xy - 4y = 4x + 3 6xy - 4 x = 4y + 3 (6y - 4 ) = 4y + 3 = 4y + 3 6y - 4 codomain Domain [ 2 3 ] f in onto function. Thus f is one-one onto function. Also, f^ -1 (x) = 4x + 3 6x - 4 .

Show that the function f in A = |R - 2 3 defined as f(x) = 4x + 3…

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Question

Show that the function f in $\text{A} = |\text{R} -\left\{\frac{2}{3}\right\}$defined as $\text{f}(\text{x}) =\frac{4\text{x} + 3}{6\text{x} - 4 }$is one-one and onto. Hence find $f^{-1}$.

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Answer

Let $x_1, x_2 \in A$
Now $\text{f}(\text{x}_{1}) = \text{f}(\text{x}_{2}) = \frac{4\text{x}_{1} + 3}{6\text{x}_{1} - 4 } = \frac{4\text{x}_{2}+ 3}{6\text{x}_{2} - 4 }$
$\Rightarrow24\text{x}_{1}\text{x}_{2} + 18 \text{x}_{2} - 16\text{x}_{1} - 12 = 24 \text{x}_{1}\text{x}_{2} + 18 \text{x}_{1} - 16 \text{x}_{2} - 12 $
$\Rightarrow - 34 \text{x}_{1} = - 34\text{x}_{2}\Rightarrow\text{x}_{1} =\text{x}_{2}$
Hence f is one-one function
For onto
Let $\text{y} =\frac{4\text{x} + 3}{6\text{x} - 4}\Rightarrow6\text{xy} - 4\text{y} = 4\text{x} + 3 $
$\Rightarrow6\text{xy} - 4 \text{x} = 4\text{y} + 3 \Rightarrow\text{x}(6\text{y} - 4 ) = 4\text{y} + 3 $
$\Rightarrow\text{x} = \frac{4\text{y} + 3 }{6\text{y} - 4}$
$\Rightarrow\forall\text{y}\in\text{ codomain}\exists\text{x}\in\text{ Domain }\bigg[\therefore\text{x}\neq\frac{2}{3}\bigg]$
$\Rightarrow$ f in onto function.
Thus f is one-one onto function.
Also, $\text{f}^{-1}\text{(x)} =\frac{4\text{x} + 3 }{6\text{x} - 4}.$