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Question: A function \(f:R \to R\) is defined as \(f\left( x \right) = {x^3} + 4\), Is it a bijection or not? ...

A function f:R→Rf:R \to R is defined as f(x)=x3+4f\left( x \right) = {x^3} + 4, Is it a bijection or not? In case it is a bijection then, find f−1(3){f^{ - 1}}\left( 3 \right)

Explanation

Solution

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

Complete step-by-step answer:
Given that,
A function f:R→Rf:R \to R is defined as f(x)=x3+4f\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(3){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(x)=f(y)f\left( x \right) = f\left( y \right)
Then,
⇒x3+4=y3+4\Rightarrow {x^3} + 4 = {y^3} + 4
⇒x3=y3\Rightarrow {x^3} = {y^3}
Taking cube root on both sides,
⇒x=y\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,
⇒f(x)=y\Rightarrow f\left( x \right) = y
Putting the value of f(x)f\left( x \right),
⇒x3+4=y\Rightarrow {x^3} + 4 = y
⇒x3=y−4\Rightarrow {x^3} = y - 4
Taking cube root on both sides, we will get
⇒x=(y−4)13\Rightarrow x = {\left( {y - 4} \right)^{\dfrac{1}{3}}}
Clearly, for all y∈R,(y−4)13y \in R,{\left( {y - 4} \right)^{\dfrac{1}{3}}} is a real number. Thus, for all y∈Ry \in R (co-domain) there exists x=(y−4)13x = {\left( {y - 4} \right)^{\dfrac{1}{3}}} in R such that f(x)=x3+4=yf\left( x \right) = {x^3} + 4 = y
Hence, f:R→Rf:R \to R is an onto function.
Here, we can see that f:R→Rf:R \to R is one-one and onto both.
Therefore, it is a bijection function.
Now,
We have to find f−1(3){f^{ - 1}}\left( 3 \right)
Let f(x)=yf\left( x \right) = y, then
⇒x3+4=y\Rightarrow {x^3} + 4 = y
⇒x3=y−4\Rightarrow {x^3} = y - 4
Taking cube root on both sides, we will get
⇒x=(y−4)13\Rightarrow x = {\left( {y - 4} \right)^{\dfrac{1}{3}}}
We can write this as:
⇒f−1(y)=(y−4)13\Rightarrow {f^{ - 1}}\left( y \right) = {\left( {y - 4} \right)^{\dfrac{1}{3}}} [∴f(x)=y⇒x=f−1(y)\therefore f\left( x \right) = y \Rightarrow x = {f^{ - 1}}\left( y \right)]
Now,
For f−1(3){f^{ - 1}}\left( 3 \right), put y = 3 in f−1(y){f^{ - 1}}\left( y \right)
⇒f−1(3)=(3−4)13\Rightarrow {f^{ - 1}}\left( 3 \right) = {\left( {3 - 4} \right)^{\dfrac{1}{3}}}
⇒f−1(3)=(−1)13\Rightarrow {f^{ - 1}}\left( 3 \right) = {\left( { - 1} \right)^{\dfrac{1}{3}}}
⇒f−1(3)=−1\Rightarrow {f^{ - 1}}\left( 3 \right) = - 1 [answer]

Note: If a function f:A→Bf:A \to B is a bijection. Then a function g:B→Ag:B \to A which associates each element y∈By \in B to a unique element x∈ax \in a such that f(x)=yf\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.