question_answer

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.