Question

How do you find the inverse of \[{x^2} + 3\] and is it a function ?

Answer 1

Hint: Here in this question, we have to find the inverse of the given function y or \[f(x)\] and check whether the inverse is a function or not. The inverse of a function is denoted by \[{f^{ - 1}}(x)\]. Here first we have to write the function in terms of x and then we have to solve for y using mathematics operations and simplification we get the required solution.

Complete step by step solution:
An inverse function or an anti-function is defined as a function, which can reverse into another function. In simple words, if any function “\[f\]” takes \[x\] to \[y\] then, the inverse of “\[f\]” will take \[y\] to \[x\]. If the function is denoted by ‘\[f\]’ or ‘\[F\]’, then the inverse function is denoted by \[{f^{ - 1}}\] or \[{F^{ - 1}}\]. i.e., If \[f\] and \[g\] are inverse functions, then \[f\left( x \right) = y\] if and only if \[g\left( y \right) = x\].

Consider the given function
\[f(x) = {x^2} + 3\]
\[\Rightarrow y = {x^2} + 3\]--------(1)
switch the \[x\]'s and the \[y\]'s means replace \[x\] as \[y\] and \[y\] as \[x\]. i.e., \[f(x)\] is a substitute for "\[y\]". Equation (1) can be written as function of \[x\]i.e.,
\[x = {y^2} + 3\]------(2)
Subtract both side by 3
\[x - 3 = {y^2} + 3 - 3\]
On simplification, we get
\[x - 3 = {y^2}\]
On rearranging
\[{y^2} = x - 3\]
Take a square root on both side
\[y = \sqrt {x - 3} \]
\[ \therefore \,\,{f^{ - 1}}\left( x \right) = \sqrt {x - 3} \]
Thus, the inverse of a function \[{x^2} + 3\] is \[\sqrt {x - 3} \].The base function of our inverse is \[\sqrt x \], which we also know it is a function.

Hence,the inverse of a function \[{x^2} + 3\] is \[\sqrt {x - 3} \].

Note: We must know about the simple arithmetic operations. To find the inverse we swap the y variable into x and simplify the equation and determine the value for y. Since the given question contains square on simplification we must know about the concept of square root. While shifting the terms we must take care of signs.