Question

What is the inverse function of $f\left( x \right)={{\left( x+1 \right)}^{2}}$ ?

Answer 1

Hint: Here in this question we have been asked to write the inverse function of the given function $f\left( x \right)={{\left( x+1 \right)}^{2}}$ . For answering this question we will first verify if the given function one to one or not because from the definition of the inverse function we know that the inverse of the function exists when the specific function is defined and has a range for each and every value in the specified domain.

Complete step-by-step answer:
Now considering from the question we have been asked to write the inverse function of the given function $f\left( x \right)={{\left( x+1 \right)}^{2}}$ .
From the basic concepts of functions we know that the definition of the inverse function say that the inverse of a function exists when the specific function is defined and has a unique value for each and every value in the specified domain that is the specified function should be one-to-one.
Now we will verify if the given function is one-to-one or not. For verifying we will substitute different values of $x$ from the set of real numbers and observe if any value is repeated or not.
Let us substitute $x=3$ then we will have $\Rightarrow f\left( x \right)={{\left( 3+1 \right)}^{2}}\Rightarrow {{4}^{2}}=16$ .
Now we will substitute $x=-5$ then we will have $\Rightarrow f\left( x \right)={{\left( -5+1 \right)}^{2}}\Rightarrow {{\left( -4 \right)}^{2}}=16$ .
Hence we can say that the given function is not one-to-one and does not have an inverse function in the domain of real numbers.

Note: This type of question should be answered by applying the concept carefully. We can define the inverse of the given function by restricting its domain from real numbers to a specific set of real numbers for the time being consider $\left[ -1,\infty \right)$ as the restricted domain if we verify in this domain the function will be one-to-one so now we will find the inverse with the following approach by assuming $f\left( x \right)={{\left( x+1 \right)}^{2}}\Rightarrow y$ and simplifying the value of $x$ in terms of $y$ which will be given as
$\begin{align}
  & \Rightarrow {{\left( y \right)}^{\frac{1}{2}}}=x+1 \\
 & \Rightarrow \sqrt{y}-1=x \\
\end{align}$ .
Now for the inverse function we need to interchange the $x$ and $y$ then the result will be the inverse function that is $\sqrt{x}-1$ .