If the function `f:RrarrR` be given by `f(x)=x^(2)+2` and `g:RrarrR` be given by `g(x)=(x)/(x-1),xne1`, find fog and gof and hence find fog (2) and gof (-3).

Answer

To solve the problem, we need to find the compositions of the functions \( f \) and \( g \), specifically \( f \circ g \) and \( g \circ f \), and then evaluate these compositions at specific points.

### Step 1: Define the functions
The functions are defined as follows:
- \( f(x) = x^2 + 2 \)
- \( g(x) = \frac{x}{x - 1} \), where \( x \neq 1 \)

### Step 2: Find \( f \circ g \)
To find \( f \circ g \), we substitute \( g(x) \) into \( f(x) \):
\[ f(g(x)) = f\left(\frac{x}{x - 1}\right) \]
Substituting \( g(x) \) into \( f(x) \):
\[ f\left(\frac{x}{x - 1}\right) = \left(\frac{x}{x - 1}\right)^2 + 2 \]
Calculating \( \left(\frac{x}{x - 1}\right)^2 \):
\[ \left(\frac{x}{x - 1}\right)^2 = \frac{x^2}{(x - 1)^2} \]
Thus,
\[ f(g(x)) = \frac{x^2}{(x - 1)^2} + 2 \]
To combine the terms:
\[ f(g(x)) = \frac{x^2 + 2(x - 1)^2}{(x - 1)^2} = \frac{x^2 + 2(x^2 - 2x + 1)}{(x - 1)^2} = \frac{3x^2 - 4x + 2}{(x - 1)^2} \]

### Step 3: Find \( g \circ f \)
Now, we find \( g \circ f \):
\[ g(f(x)) = g(x^2 + 2) \]
Substituting \( f(x) \) into \( g(x) \):
\[ g(x^2 + 2) = \frac{x^2 + 2}{(x^2 + 2) - 1} = \frac{x^2 + 2}{x^2 + 1} \]

### Step 4: Evaluate \( f \circ g(2) \)
Now we evaluate \( f \circ g(2) \):
\[ f(g(2)) = f\left(g(2)\right) \]
First, calculate \( g(2) \):
\[ g(2) = \frac{2}{2 - 1} = 2 \]
Now substitute into \( f \):
\[ f(g(2)) = f(2) = 2^2 + 2 = 4 + 2 = 6 \]

### Step 5: Evaluate \( g \circ f(-3) \)
Now we evaluate \( g(f(-3)) \):
First, calculate \( f(-3) \):
\[ f(-3) = (-3)^2 + 2 = 9 + 2 = 11 \]
Now substitute into \( g \):
\[ g(f(-3)) = g(11) = \frac{11}{11 - 1} = \frac{11}{10} \]

### Final Results
- \( f \circ g(2) = 6 \)
- \( g \circ f(-3) = \frac{11}{10} \)