Question

Given a function, y=f(x), and its inverse, f-1(x)a. −1[()] = b. −1[()] = c. −1/() = 1d. −1+()=e. None of the above

Answer 1

$f^{-1}\lbrack(f(x)\rbrack\text{ = x (option A)}$

Explanation:$\begin{gathered} \text{y = }f(x) \\ \text{f}^{-1\text{ }}(x)\text{ = inverse of f(x)} \\ \text{if x = 4},\text{ and f(x)= 10} \\ \text{then f}^{-1\text{ }}(10)\text{ = }4 \end{gathered}$

This shows that when we apply a function f(x) and its inverse f^-1(x) , we would get back the initial input which is x

$\begin{gathered} when\text{ x = input and f(x) = output} \\ The\text{ inverse of the function of x will give back x as the result} \\ In\text{ other words (symbolically):} \\ f^{-1}(f(x))\text{ = x} \\ we\text{ can also have:} \\ f\lbrack(f^{-1}(x)\rbrack\text{ = x} \end{gathered}$$\begin{gathered} \text{Hence, the correct answer:} \\ f^{-1}\lbrack(f(x)\rbrack\text{ = x (option A)} \end{gathered}$