Question

A function $f:R \to R$ is defined as $f\left( x \right) = {x^3} + 4$, Is it a bijection or not? In case it is a bijection then, find ${f^{ - 1}}\left( 3 \right)$

Answer 1

Hint: To solve this question, we have to remember that a function $f:A \to B$ is a bijection if it is one-one as well as onto.
(i) one-one i.e. $f\left( x \right) = f\left( y \right) \Rightarrow x = y\forall x,y \in A$
(ii) onto i.e. for all $y \in B$, there exists $X \in A$ such that $f\left( x \right) = y$

Complete step-by-step answer:
Given that,
A function $f:R \to R$ is defined as $f\left( x \right) = {x^3} + 4$
 We have to find out whether the given function is a bijection or not.
If so, then find ${f^{ - 1}}\left( 3 \right)$
First, let us check for one-one.
Let x and y be two arbitrary elements of R (domain of f) such that $f\left( x \right) = f\left( y \right)$
Then,
$ \Rightarrow {x^3} + 4 = {y^3} + 4$
$ \Rightarrow {x^3} = {y^3}$
Taking cube root on both sides,
$ \Rightarrow x = y$
Hence, f is one-one function from R to itself.
Now, we will check for onto.
Let y be an arbitrary element of R.
Then,
$ \Rightarrow f\left( x \right) = y$
Putting the value of $f\left( x \right)$,
$ \Rightarrow {x^3} + 4 = y$
$ \Rightarrow {x^3} = y - 4$
Taking cube root on both sides, we will get
$ \Rightarrow x = {\left( {y - 4} \right)^{\dfrac{1}{3}}}$
Clearly, for all $y \in R,{\left( {y - 4} \right)^{\dfrac{1}{3}}}$ is a real number. Thus, for all $y \in R$ (co-domain) there exists $x = {\left( {y - 4} \right)^{\dfrac{1}{3}}}$ in R such that $f\left( x \right) = {x^3} + 4 = y$
Hence, $f:R \to R$ is an onto function.
Here, we can see that $f:R \to R$ is one-one and onto both.
Therefore, it is a bijection function.
Now,
We have to find ${f^{ - 1}}\left( 3 \right)$
Let $f\left( x \right) = y$, then
$ \Rightarrow {x^3} + 4 = y$
$ \Rightarrow {x^3} = y - 4$
Taking cube root on both sides, we will get
$ \Rightarrow x = {\left( {y - 4} \right)^{\dfrac{1}{3}}}$
We can write this as:
$ \Rightarrow {f^{ - 1}}\left( y \right) = {\left( {y - 4} \right)^{\dfrac{1}{3}}}$ [$\therefore f\left( x \right) = y \Rightarrow x = {f^{ - 1}}\left( y \right)$]
Now,
For ${f^{ - 1}}\left( 3 \right)$, put y = 3 in ${f^{ - 1}}\left( y \right)$
$ \Rightarrow {f^{ - 1}}\left( 3 \right) = {\left( {3 - 4} \right)^{\dfrac{1}{3}}}$
$ \Rightarrow {f^{ - 1}}\left( 3 \right) = {\left( { - 1} \right)^{\dfrac{1}{3}}}$
$ \Rightarrow {f^{ - 1}}\left( 3 \right) = - 1$ [answer]

Note: If a function $f:A \to B$ is a bijection. Then a function $g:B \to A$ which associates each element $y \in B$ to a unique element $x \in a$ such that $f\left( x \right) = y$ is called the inverse of f. For finding the inverse of a function, that function must be a bijection function, otherwise the inverse of that function does not exist.