Prove that function `f:RrarrR,f(x)=(3-2x)/(7)` is one-one onto. Also, find `f^(-1)`.

Answer

To prove that the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = \frac{3 - 2x}{7} \) is one-one and onto, and to find its inverse \( f^{-1}(x) \), we will follow these steps:

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

A function is one-one (injective) if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).

Assume \( f(x_1) = f(x_2) \):
\[ \frac{3 - 2x_1}{7} = \frac{3 - 2x_2}{7} \]
Multiply both sides by 7:
\[ 3 - 2x_1 = 3 - 2x_2 \]
Subtract 3 from both sides:
\[ -2x_1 = -2x_2 \]
Divide by -2:
\[ x_1 = x_2 \]
Since \( x_1 = x_2 \), the function \( f \) is one-one.

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

A function is onto (surjective) if for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).

Let \( y = f(x) = \frac{3 - 2x}{7} \). We need to solve for \( x \):
\[ y = \frac{3 - 2x}{7} \]
Multiply both sides by 7:
\[ 7y = 3 - 2x \]
Rearranging gives:
\[ 2x = 3 - 7y \]
Dividing by 2:
\[ x = \frac{3 - 7y}{2} \]
Since \( y \) can be any real number, we can find an \( x \) for every \( y \). Hence, \( f \) is onto.

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

From the previous step, we derived:
\[ x = \frac{3 - 7y}{2} \]
To express this in terms of \( y \), we replace \( y \) with \( x \):
\[ f^{-1}(x) = \frac{3 - 7x}{2} \]

### Conclusion

The function \( f(x) = \frac{3 - 2x}{7} \) is one-one and onto, and its inverse is:
\[ f^{-1}(x) = \frac{3 - 7x}{2} \]