If f : `ArarrAandA=R-{8/(5)}`, show that the function f(x) =`(8x+3)/(5x-8)` is one -one onto. Hence find `f^(-1)`.

Answer

To show that the function \( f(x) = \frac{8x + 3}{5x - 8} \) is one-one and onto, and to find its inverse, we will follow these steps:

### Step 1: Proving that \( f \) is a one-one function

To prove that \( f \) is one-one, we need to show that if \( f(x_1) = f(x_2) \), then \( x_1 = x_2 \).

1. Assume \( f(x_1) = f(x_2) \):
\[ \frac{8x_1 + 3}{5x_1 - 8} = \frac{8x_2 + 3}{5x_2 - 8} \]

2. Cross-multiply:
\[ (8x_1 + 3)(5x_2 - 8) = (8x_2 + 3)(5x_1 - 8) \]

3. Expanding both sides:
\[ 40x_1x_2 - 64x_1 + 15x_2 - 24 = 40x_1x_2 - 64x_2 + 15x_1 - 24 \]

4. Cancel \( 40x_1x_2 \) and \( -24 \) from both sides:
\[ -64x_1 + 15x_2 = -64x_2 + 15x_1 \]

5. Rearranging gives:
\[ 64x_2 - x_1 + 15x_2 - x_1 = 0 \]
\[ 79x_2 - x_1 = 0 \]

6. Thus, we have:
\[ x_1 = x_2 \]

Since \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \), the function \( f \) is one-one.

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

To show that \( f \) is onto, we need to show that for every \( y \in \mathbb{R} \setminus \{ \frac{8}{5} \} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).

1. Set \( f(x) = y \):
\[ y = \frac{8x + 3}{5x - 8} \]

2. Rearranging gives:
\[ y(5x - 8) = 8x + 3 \]

3. Expanding and rearranging:
\[ 5yx - 8y = 8x + 3 \]
\[ 5yx - 8x = 8y + 3 \]
\[ x(5y - 8) = 8y + 3 \]

4. Solving for \( x \):
\[ x = \frac{8y + 3}{5y - 8} \]

5. The function is defined for all \( y \) except \( y = \frac{8}{5} \) (as it would make the denominator zero).

Since we can find an \( x \) for every \( y \) in \( \mathbb{R} \setminus \{ \frac{8}{5} \} \), the function \( f \) is onto.

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

From the previous steps, we derived:
\[ x = \frac{8y + 3}{5y - 8} \]

Now, to express \( y \) in terms of \( x \):
1. Swap \( x \) and \( y \):
\[ y = \frac{8x + 3}{5x - 8} \]

Thus, the inverse function is:
\[ f^{-1}(x) = \frac{8x + 3}{5x - 8} \]

### Summary of Results
- The function \( f(x) = \frac{8x + 3}{5x - 8} \) is one-one and onto.
- The inverse function is \( f^{-1}(x) = \frac{8x + 3}{5x - 8} \).