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Home  >>  CBSE XII  >>  Math  >>  Relations and Functions
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Let \(f : R - \{ - \frac {4} {3} \} \to R \) be a function defined as \(f(x)= \frac {4x} {3x+4} \) .The inverse of \(f\) is the map \(g\): Range \(f \to R - \{ - \frac {4} {3} \} \) given by

 \begin{array}{1 1}(a)\;\;\;\;\; g(y)= \frac {3y} {3-4y} & (b)\;\;\;\;\; g(y)= \frac {4y} {4-3y}\\(c)\;\;\;\;\; g(y)= \frac {4y} {3-4y} & (d)\;\;\;\;\; g(y)= \frac {3y} {4-3y}\end{array}

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Toolbox:
  • A function $g$ is called inverse of $f:x \to y$, then exists $g:y \to x$ such that $ gof=I_x\;and\; fog=I_y$, where $I_x, I_y$ are identify functions.
  • Given two functions $f:A \to B $ and $g:B \to C$, then composition of $f$ and $g$, $gof:A \to C$ by $ gof (x)=g(f(x))\;for\; all \;x \in A$
Given a function $f:R-${$\frac{-4}{3}$}$\to R$ defined by $f(x)=\large \frac{4x}{3x+4}$.
$\textbf {Step 1: To calculate the inverse of } f, \textbf {we must first define} \;g(y):$
Let $y = f(x) =\large \frac{4x}{3x+4}$.
$\Rightarrow 3xy + 4y = 4x \rightarrow x =\large \frac{4y}{4-3y}$
Let us now define a function $g:$Range $f \to R-\{\frac{-4}{3}\}$ given by $g(y)=\large \frac{4y}{4-3y}$
$\textbf {Step 2: Calculating} \;gof$:
We know that $(gof) (x)=g(f(x))$
$\Rightarrow gof =g(\frac{4x}{3x+4})$
$\Rightarrow gof =\Large\frac{4(\frac{4x}{3x+4})}{4-3(\frac{4x}{3x+4})}$
$\Rightarrow gof =\Large \frac{16x}{12x+16-12x}=\frac{16x}{16}$ = $x$
$\textbf {Step 3: Calculating} \;fog$:
We know that $(fog) (y)=f(g(y))$
$\Rightarrow fog$ $=f(\frac{4y}{4-3y})$
$\Rightarrow fog$ $=\Large\frac{4(\frac{4y}{4-3y})}{3(\frac{4y}{4-3y})+4}$
$\Rightarrow fog$ $=\Large \frac{16y}{12y+16-12y}=\frac{16y}{16}$ = $y$
$\textbf {Step 4: Calculating} \;\;f^{-}\; \textbf {from}\; \;gof = fog$
$gof=I_{R-\{\frac{4}{3}\}}$ and $fog=I_{Range f}$
Therefore $ f^{-1}=g$ defined by $g:Range f \to R-\{\frac{4}{3}\}$ where $g(y)=\large \frac{4y}{4-3y}$
Therefore $(B) \; g(y)=\large \frac{4y}{4-3y}$ is the correct answer.
answered Feb 26, 2013 by meena.p
edited Jul 17, 2014 by balaji.thirumalai
 

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