We are given two functions, $f(x) = x^2 + 1$ and $g(x) = 5 - 3x$, defined on the set of real numbers. (i) We need to find the domain of the inverse function $f^{-1}$. (ii) We need to find the value of the inverse function $g^{-1}(2)$.

1. Problem Description

We are given two functions, $f(x) = x^2 + 1$ and $g(x) = 5 - 3x$, defined on the set of real numbers.

(i) We need to find the domain of the inverse function $f^{-1}$.

(ii) We need to find the value of the inverse function $g^{-1}(2)$.

2. Solution Steps

(i) To find the domain of $f^{-1}$, we need to find the range of $f(x)$.

Since $x$ is a real number, $x^2$ is always non-negative, i.e., $x^2 \ge 0$.

Therefore, $f(x) = x^2 + 1 \ge 0 + 1 = 1$.

Thus, the range of $f(x)$ is $[1, \infty)$.

The domain of $f^{-1}$ is equal to the range of $f(x)$.

Hence, the domain of $f^{-1}$ is $[1, \infty)$.

(ii) To find $g^{-1}(2)$, we first need to find the expression for $g^{-1}(x)$.

Let $y = g(x) = 5 - 3x$.

To find the inverse, we swap $x$ and $y$ and solve for $y$.

So, $x = 5 - 3y$.

$3y = 5 - x$

$y = \frac{5 - x}{3}$

Therefore, $g^{-1}(x) = \frac{5 - x}{3}$.

Now, we substitute $x = 2$ into the expression for $g^{-1}(x)$:

$g^{-1}(2) = \frac{5 - 2}{3} = \frac{3}{3} = 1$.

3. Final Answer

(i) The domain of $f^{-1}$ is $[1, \infty)$.

(ii) $g^{-1}(2) = 1$.