(a) If `f:RrarrR` defined by `f(x)=(3x+5)/(2)` is an invertible function, find `f^(-1)`.
Show that `f:RrarrR` defined by `f(x)=(4x-3)/(5),x inR` is invertible function and find `f^(-1)`.

Answer

To solve the given problems, we need to determine if the functions are invertible and then find their inverses if they are. A function is invertible if it is both one-to-one (injective) and onto (surjective).

### Part (a): Finding the inverse of \( f(x) = \frac{3x + 5}{2} \)

1. **Check if the function is one-to-one:**
- Assume \( f(x_1) = f(x_2) \).
- This means \( \frac{3x_1 + 5}{2} = \frac{3x_2 + 5}{2} \).
- Multiply both sides by 2:
\[ 3x_1 + 5 = 3x_2 + 5 \]
- Subtract 5 from both sides:
\[ 3x_1 = 3x_2 \]
- Divide by 3:
\[ x_1 = x_2 \]
- Since \( x_1 = x_2 \), the function is one-to-one.

2. **Check if the function is onto:**
- Let \( y \) be any real number. We need to find \( x \) such that \( f(x) = y \):
\[ \frac{3x + 5}{2} = y \]
- Multiply both sides by 2:
\[ 3x + 5 = 2y \]
- Subtract 5 from both sides:
\[ 3x = 2y - 5 \]
- Divide by 3:
\[ x = \frac{2y - 5}{3} \]
- Since \( x \) can be expressed in terms of \( y \), the function is onto.

3. **Conclusion:**
- Since \( f(x) \) is both one-to-one and onto, it is invertible.

4. **Find the inverse:**
- From the equation \( y = \frac{3x + 5}{2} \), we already found:
\[ x = \frac{2y - 5}{3} \]
- Thus, the inverse function is:
\[ f^{-1}(y) = \frac{2y - 5}{3} \]

### Part (b): Finding the inverse of \( f(x) = \frac{4x - 3}{5} \)

1. **Check if the function is one-to-one:**
- Assume \( f(x_1) = f(x_2) \).
- This means \( \frac{4x_1 - 3}{5} = \frac{4x_2 - 3}{5} \).
- Multiply both sides by 5:
\[ 4x_1 - 3 = 4x_2 - 3 \]
- Add 3 to both sides:
\[ 4x_1 = 4x_2 \]
- Divide by 4:
\[ x_1 = x_2 \]
- Since \( x_1 = x_2 \), the function is one-to-one.

2. **Check if the function is onto:**
- Let \( y \) be any real number. We need to find \( x \) such that \( f(x) = y \):
\[ \frac{4x - 3}{5} = y \]
- Multiply both sides by 5:
\[ 4x - 3 = 5y \]
- Add 3 to both sides:
\[ 4x = 5y + 3 \]
- Divide by 4:
\[ x = \frac{5y + 3}{4} \]
- Since \( x \) can be expressed in terms of \( y \), the function is onto.

3. **Conclusion:**
- Since \( f(x) \) is both one-to-one and onto, it is invertible.

4. **Find the inverse:**
- From the equation \( y = \frac{4x - 3}{5} \), we already found:
\[ x = \frac{5y + 3}{4} \]
- Thus, the inverse function is:
\[ f^{-1}(y) = \frac{5y + 3}{4} \]

### Summary of Results:
- For \( f(x) = \frac{3x + 5}{2} \), the inverse is \( f^{-1}(y) = \frac{2y - 5}{3} \).
- For \( f(x) = \frac{4x - 3}{5} \), the inverse is \( f^{-1}(y) = \frac{5y + 3}{4} \).