If the function f and g are defined as `f(x)=e^(x)` and g(x)=3x-2, where `f:R rarr R` and `g:R rarr R`, find the function fog and gof. Also, find the domain of `(fog)^(-1) " and " (gof)^(-1)`.

Answer

To solve the problem, we need to find the compositions of the functions \( f \) and \( g \), denoted as \( f \circ g \) and \( g \circ f \), and then determine the domains of their inverses.

### Step 1: Define the Functions
The functions are defined as:
- \( f(x) = e^x \)
- \( g(x) = 3x - 2 \)

### Step 2: Find \( f \circ g \)
To find \( f \circ g \), we substitute \( g(x) \) into \( f(x) \):
\[ f(g(x)) = f(3x - 2) = e^{3x - 2} \]
Thus,
\[ f \circ g = e^{3x - 2} \]

### Step 3: Find \( g \circ f \)
To find \( g \circ f \), we substitute \( f(x) \) into \( g(x) \):
\[ g(f(x)) = g(e^x) = 3(e^x) - 2 = 3e^x - 2 \]
Thus,
\[ g \circ f = 3e^x - 2 \]

### Step 4: Find the Inverse of \( f \circ g \)
Let \( y = f \circ g(x) = e^{3x - 2} \). To find the inverse, we solve for \( x \):
1. Take the natural logarithm of both sides:
\[ \ln(y) = 3x - 2 \]
2. Rearranging gives:
\[ 3x = \ln(y) + 2 \]
3. Dividing by 3:
\[ x = \frac{\ln(y) + 2}{3} \]
Thus, the inverse function is:
\[ (f \circ g)^{-1}(y) = \frac{\ln(y) + 2}{3} \]

### Step 5: Find the Domain of \( (f \circ g)^{-1} \)
The domain of \( (f \circ g)^{-1} \) is determined by the requirement that \( y > 0 \) (since the logarithm is defined only for positive values):
\[ y > 0 \implies \text{Domain: } (0, \infty) \]

### Step 6: Find the Inverse of \( g \circ f \)
Let \( y = g \circ f(x) = 3e^x - 2 \). To find the inverse, we solve for \( x \):
1. Rearranging gives:
\[ 3e^x = y + 2 \]
2. Dividing by 3:
\[ e^x = \frac{y + 2}{3} \]
3. Taking the natural logarithm:
\[ x = \ln\left(\frac{y + 2}{3}\right) \]
Thus, the inverse function is:
\[ (g \circ f)^{-1}(y) = \ln\left(\frac{y + 2}{3}\right) \]

### Step 7: Find the Domain of \( (g \circ f)^{-1} \)
The domain of \( (g \circ f)^{-1} \) is determined by the requirement that \( \frac{y + 2}{3} > 0 \):
\[ y + 2 > 0 \implies y > -2 \implies \text{Domain: } (-2, \infty) \]

### Summary of Results
- \( f \circ g = e^{3x - 2} \)
- \( g \circ f = 3e^x - 2 \)
- Domain of \( (f \circ g)^{-1} \): \( (0, \infty) \)
- Domain of \( (g \circ f)^{-1} \): \( (-2, \infty) \)