If \[f:R\to R\] is a function defined by \[f(x)=2{{x}^{3}}-5,\] then show that the function f is a objective function. |
OR |
Consider \[f:R\to R\] given by \[f(x)=4x+3.\]Show that f is invertible and find the inverse off. |
Answer:
Given, \[f:R\to R\] defined by \[f(x)=2{{x}^{3}}-5\] For one-one (injective) Let \[f({{x}_{1}})=f({{x}_{2}}),\,\,\forall {{x}_{1}},\,{{x}_{2}}\in R\] \[\Rightarrow \] \[2x_{1}^{3}-5=2x_{2}^{3}-5\] \[\Rightarrow \] \[2x_{1}^{3}=2x_{2}^{3}\] \[\Rightarrow \] \[x_{1}^{3}=x_{2}^{3}\] \[\Rightarrow \] \[{{x}_{1}}={{x}_{2}}\] Thus, \[f({{x}_{1}})=f({{x}_{2}})\] \[\Rightarrow \] \[{{x}_{1}}={{x}_{2}}\] So, f is one-one (injective) For onto (surjective) Let y be an arbitrary element of R (codomain), then \[f(x)=y\] \[\Rightarrow \] \[2{{x}^{3}}-5=y\] \[\Rightarrow \] \[2{{x}^{3}}=y+5\] \[\Rightarrow \] \[{{x}^{3}}=\frac{y+5}{2}\] \[\Rightarrow \] \[x={{\left( \frac{y+5}{2} \right)}^{1/3}}\] Clearly, \[x\in R\] (domain), \[\forall \,\,y\in R\] (codomain). Thus, for each \[y\in R\] (codomain) there exists \[x={{\left( \frac{y+5}{2} \right)}^{1/3}}\in R\] (domain) such that \[f(x)=f\left[ {{\left( \frac{y+5}{2} \right)}^{1/3}} \right]=2{{\left\{ {{\left( \frac{y+5}{2} \right)}^{1/3}} \right\}}^{3}}-5\] \[=2\left[ \left( \frac{y+5}{2} \right) \right]-5=y+5-5=y\] This shows that every element in the codomain has its pre-image in the domain. So, f is onto (or f s surjective). Thus, f is both one-one and onto (or both injective and subjective). Hence, f is bijective. Hence proved. OR Given, \[f:R\to R\] defined by \[f(x)=4x+3\] For one-one Let \[{{x}_{1}}\] and \[{{x}_{2}}\] be two arbitrary elements of R, Then \[f({{x}_{1}})=f({{x}_{2}})\] \[\Rightarrow \] \[4{{x}_{1}}+3=4{{x}_{2}}+3\] \[\Rightarrow \] \[4{{x}_{1}}=4{{x}_{2}}\] \[\Rightarrow \] \[{{x}_{1}}={{x}_{2}}\] Thus, \[f({{x}_{1}})=f({{x}_{2}})\] \[\Rightarrow {{x}_{1}}={{x}_{2}},.\] \[\forall {{x}_{1}},\,\,{{x}_{2}}\in R\] So, f is one-one. For onto Let y be an arbitrary element of R (codomain), then \[y=f(x)\,\,\,\,\Rightarrow \,\,\,\,y=4x+3\] \[\Rightarrow \] \[4x=y-3\] \[\Rightarrow \] \[x=\frac{y-3}{4}\] Clearly, \[x=\frac{y-3}{4}\in R,\] \[\forall \,y\in R.\] Thus, for each \[y\in R\] (codomain), there exists \[x=\frac{y-3}{4}\in R\] such that f(x) = y So, f is onto. Hence, f is one-one and onto, so f is invertible. Then, there exists \[g:R\to R\] such that \[gof={{l}_{R}}\] \[\therefore \] \[gof(x)=x,\] \[\forall \times \in R\] \[\Rightarrow \] \[g[f(x)]=x,\] \[\forall \times \in R\] \[\Rightarrow \] \[g[4x+3]=x\] where, \[y=f(x)=4x+3\] \[\Rightarrow \] \[g(y)=\frac{y-3}{4}={{f}^{-1}}(y)\] Thus, \[{{f}^{-1}}:R\to R\] is defined as \[{{f}^{-1}}(x)=\frac{x-3}{4},\] \[\forall \times \in R.\] Hence proved.
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