If \[f:R-\{2\}\to R-\{3\}\] is defined by |
\[f(x)=\frac{3x+1}{x-2},\] where R is the set of real numbers, |
show that f is invertible and hence find the value of \[{{f}^{-1}}.\] |
OR |
Let \[f:N\to R\] be a function defined as \[f(x)=4{{x}^{2}}+12x+15.\] Show that \[f:N\to \] range f is invertible. Find the inverse of \[{{f}^{-1}}\]. |
Answer:
Given, \[f\,(x)=\frac{3x+1}{x-2}\]and \[f\,(x);R-\{z\}\to R-\{z\}\] For one-one Let\[f\,({{x}_{1}})=f\,({{x}_{2}}),\]for some \[{{x}_{1}},\]\[{{x}_{2}}\in A\] \[\Rightarrow \] \[\frac{3{{x}_{1}}+1}{{{x}_{1}}-2}=\frac{3{{x}_{2}}+1}{{{x}_{2}}-2}\] \[\Rightarrow \] \[(3{{x}_{1}}+1)\,({{x}_{2}}-2)=(3{{x}_{2}}+1)\,({{x}_{1}}-2)\] \[\Rightarrow \]\[3{{x}_{1}}{{x}_{2}}-6{{x}_{1}}+{{x}_{2}}-2=3{{x}_{1}}{{x}_{2}}-6{{x}_{2}}+{{x}_{1}}-2\] \[\Rightarrow \] \[-\,6{{x}_{1}}+{{x}_{2}}=-\,6{{x}_{2}}+{{x}_{1}}\] \[\Rightarrow \] \[-\,6{{x}_{1}}-{{x}_{1}}=-\,6{{x}_{2}}-{{x}_{2}}\] \[\Rightarrow \] \[-\,7{{x}_{1}}-7{{x}_{2}}\] \[\Rightarrow \] \[{{x}_{1}}={{x}_{2}}\] So, \[f\,(x)\] is one-one function. For onto Let \[y=\frac{3x+1}{x-2},\]then \[\Rightarrow \] \[xy-2y=3x+1\] \[\Rightarrow \] \[xy-3x=2y+1\] \[\Rightarrow \] \[x\,(y-3)=2y+1\] \[\Rightarrow \] \[x=\frac{2y+1}{y-3}\] Since, \[x\in R-\{2\},\,\,\forall \in R-\{3\}\]. So, range of\[f\,(x)=R-\{z\}\] \[\therefore \] Range = Codomain So, \[f\,(x)\] is onto function. (1) Also, from Eq. (i), we get \[{{f}^{-\,1}}(y)=\frac{2y+1}{y-3}\] \[[\because x={{f}^{-\,1}}(y)]\] Or \[{{f}^{-\,1}}(x)=\frac{2x+1}{x-3}\] OR We have a mapping \[f:N\to R\] defined as \[f\,(x)=4{{x}^{2}}+12x+15\] To show \[f:N\to R\] range \[(f)\] is invertible. For this it is sufficient to prove that f is one-one. [\[\because \]range \[(f)\]= codomain \[(f)\Rightarrow f\] is onto] For one-one Let \[{{x}_{1}},\]\[{{x}_{2}}\in N\]such that \[f\,({{x}_{1}})=f\,({{x}_{2}})\] Then, \[4x_{1}^{2}+12{{x}_{1}}+15=4x_{2}^{2}+12{{x}_{2}}+15\] \[\Rightarrow \] \[4\,(x_{1}^{2}-x_{2}^{2})=12\,({{x}_{2}}-{{x}_{1}})\] \[\Rightarrow \] \[(4\,({{x}_{1}}+{{x}_{2}})+12)\,\,({{x}_{1}}-{{x}_{2}})=0\] \[\Rightarrow \] \[{{x}_{1}}-{{x}_{2}}=0\] \[[\because 4{{x}_{1}}+4{{x}_{2}}+12\ne 0\,\,as\,\,{{x}_{1}},\,\,{{x}_{2}}\in N]\] \[\Rightarrow \] \[{{x}_{1}}={{x}_{2}}\] \[\Rightarrow \] f is one-one. Hence \[f:N\to \] range \[(f)\] is invertible. Now, to find inverse of f , let \[y=f\,(x)\Rightarrow y=4{{x}^{2}}+12x+15\] \[\therefore \] \[4{{x}^{2}}+12x+15-y=0\] Now, \[x=\frac{-\,12\,\,\pm \sqrt{144-4\,(4)\,\,(15-y)}}{8}\] [by Sridharacharya?s formula] \[=\frac{-\,12\,\,\pm \,\,4\sqrt{9-(15-y)}}{8}\] \[=\frac{-\,3\pm \sqrt{y-6}}{2}\] But \[x\ne \frac{-\,3-\sqrt{y-6}}{2}\] \[[\because x\in N]\] \[\therefore \] \[x=\frac{-\,3+\sqrt{y-6}}{2}\] \[\Rightarrow {{f}^{-\,1}}(y)=\frac{-\,3+\sqrt{y-6}}{2}\] So, \[{{f}^{-\,1}}:\]range \[(f)\to N\]defined as\[{{f}^{-\,1}}(x)=\frac{-\,3+\sqrt{x-6}}{2}\].
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