Show that function `f:A rarr B` defined as `f(x)=(3x+4)/(5x-7)` where `A=R-{7/5}, B=R-{3/5}` is invertible and hence find `f^(-1)`.

Answer

To show that the function \( f: A \to B \) defined by

\[ f(x) = \frac{3x + 4}{5x - 7} \]

where \( A = \mathbb{R} - \left\{ \frac{7}{5} \right\} \) and \( B = \mathbb{R} - \left\{ \frac{3}{5} \right\} \) is invertible, we need to prove that it is a bijective function (both one-to-one and onto). After proving that it is invertible, we will find the inverse function \( f^{-1} \).

### Step 1: Prove that \( f \) is one-to-one

To show that \( f \) is one-to-one, we assume that \( f(a) = f(b) \) for some \( a, b \in A \). This means:

\[ \frac{3a + 4}{5a - 7} = \frac{3b + 4}{5b - 7} \]

Cross-multiplying gives:

\[ (3a + 4)(5b - 7) = (3b + 4)(5a - 7) \]

Expanding both sides:

\[ 15ab - 21a + 20b - 28 = 15ab - 21b + 20a - 28 \]

Cancelling \( 15ab \) and \( -28 \) from both sides:

\[ -21a + 20b = -21b + 20a \]

Rearranging gives:

\[ -21a + 21b = 20a - 20b \]

Combining like terms:

\[ 41b = 41a \]

Thus, we conclude that:

\[ a = b \]

This shows that \( f \) is one-to-one.

### Step 2: Prove that \( f \) is onto

To show that \( f \) is onto, we need to show that for every \( y \in B \), there exists an \( x \in A \) such that \( f(x) = y \).

Let \( y \in B \). We set:

\[ y = \frac{3x + 4}{5x - 7} \]

Cross-multiplying gives:

\[ y(5x - 7) = 3x + 4 \]

Expanding this:

\[ 5xy - 7y = 3x + 4 \]

Rearranging terms gives:

\[ 5xy - 3x = 7y + 4 \]

Factoring out \( x \):

\[ x(5y - 3) = 7y + 4 \]

Thus, we can solve for \( x \):

\[ x = \frac{7y + 4}{5y - 3} \]

Since \( y \neq \frac{3}{5} \) (as per the definition of \( B \)), \( 5y - 3 \neq 0 \). Therefore, \( x \) is well-defined and belongs to \( A \).

This shows that \( f \) is onto.

### Conclusion

Since \( f \) is both one-to-one and onto, we conclude that \( f \) is bijective, and thus invertible.

### Step 3: Find the inverse function \( f^{-1} \)

From our calculation in the onto step, we found that:

\[ f^{-1}(y) = \frac{7y + 4}{5y - 3} \]

To express it in terms of \( x \):

\[ f^{-1}(x) = \frac{7x + 4}{5x - 3} \]

### Final Answer

The function \( f \) is invertible, and the inverse function is:

\[ f^{-1}(x) = \frac{7x + 4}{5x - 3} \]