Consider f : R - -4 3 R - 4 3 given by f(x) = 4x + 3 3x + 4 . Sho… - Vidyadip

Question

Consider $\text{f : R} - \left\{-\frac{4}{3}\right\} \rightarrow \text{R} - \left\{\frac{4}{3}\right\} \text{given by f(x)} = \frac{\text{4x + 3}}{\text{3x + 4}}.$ Show that f is bijective. Find the inverse of f and hence find $f ^{–1}(0)$ and x such that $f ^{–1}(x) = 2$.

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Answer

$\text{Let } \text{x}_{1}, \text{x}_{2} \in \text{R} - \left\{-\frac{4}{3}\right\} \text{and } \text{f(x}_{1}) = \text{f(x}_{2})$
$\Rightarrow \frac{\text{4x}_{1} + 3}{\text{3x}_{1} + 4} = \frac{\text{4x}_{2} + 3}{\text{3x}_{2} + 4} \Rightarrow \text{(4x}_{1} + 3) \text{(3x}_{2} + 4) = \text{(3x}_{1} + 4) \text{(4x}_{2} + 3)$
$\Rightarrow \text{12x}_{1} \text{x}_{2} + \text{16x}_{1} + \text{9x}_{2} + 12 = 12_{1} \text{x}_{2} + 16\text{x}_{2} + 9\text{x}_{1} + 12$
$\Rightarrow 16(\text{x}_{1} - \text{x}_{2}) - 9 \text{(x}_{1} - \text{x}_{2}) = 0 \Rightarrow \text{x}_{1} - \text{x}_{2} = 0 \Rightarrow \text{x}_{1} = \text{x}_{2}$
Hence f is a 1–1 function
$\text{Let} \text{ y} = \frac{\text{4x + 3}}{\text{3x + 4}}, \text{for y} \in \text{R} - \left\{\frac{4}{3}\right\}$
$\text{3xy + 4y = 4x + 3} \Rightarrow \text{4x – 3xy = 4y – 3}$
$\Rightarrow \text{x} = \frac{\text{4y - 3}}{\text{4 - 3y}} \therefore \forall \text{ y} \in \text{R} - \left\{\frac{4}{3}\right\}, \text{x} \in \text{R} - \left\{-\frac{4}{3}\right\}$
Hence f is ONTO and so bijective
$\text{and } \text{f}^{-1} \text{(y)} = \frac{\text{4y - 3}}{\text{4 - 3y}} ; \text{y} \in \text{R} - \left\{\frac{4}{3}\right\}$
$\text{f}^{-1} (0) = - \frac{3}{4}$
$\text{and } \text{f}^{-1} \text{(x)} = 2 \Rightarrow \frac{\text{4x - 3}}{\text{4 - 3x}} = 2$
$\Rightarrow \text{4x - 3 = 8 - 6x}$
$\Rightarrow \text{10x} = 11 \Rightarrow \text{x} = \frac{11}{10}$

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