Question

Consider the function f(x)=(x+2)^2 for the Domain [-2, +infinity). find f^-1(x), where f^-1 is the inverse of f. Also state the domain of f^-1 in internal notation.

Answer 1

SOLUTION

Step 1 :

In this question, we are meant to consider the function,

$f(x)=(x+2)^2\text{ for the domain }\lbrack\text{ 2 , }\infty\text{ })$

Step 2 :

We are meant to find the inverse of f as follows:

$\begin{gathered} \text{Let y = f ( x )} \\ y=(x+2)^2 \\ \text{Taking the square - root of both sides, we have that :} \\ \sqrt[\square]{y}\text{ = x + 2} \\ \text{Simplifying further, we have that:} \\ x\text{ = }\sqrt[]{y\text{ }}\text{ - 2} \\ f^{-1}(\text{ x) = }\sqrt[]{x\text{ }}\text{ - 2 = y} \\ \text{Therefore y = }\sqrt[]{x\text{ }}\text{ - 2} \end{gathered}$

Step 3 :

For the second part of the question,

We need to get the domain of

$f^{-1}(\text{ x)}$$\begin{gathered} \text{The domain of f}^{-1}(\text{ x ) = x }\ge\text{ 0 } \\ =\text{ }\lbrack\text{ 0 , }\infty\text{ ) ---- Interval notation.} \end{gathered}$