If the function `f(x)=sqrt(2x-3)` is invertible then, find its inverse. Hence prove that `(fof^(-1)) (x)=x`.

Answer

To find the inverse of the function \( f(x) = \sqrt{2x - 3} \) and prove that \( (f \circ f^{-1})(x) = x \), we can follow these steps:

### Step 1: Write the function
We start with the function:
\[ f(x) = \sqrt{2x - 3} \]

### Step 2: Set \( y = f(x) \)
Let \( y = f(x) \). This gives us:
\[ y = \sqrt{2x - 3} \]

### Step 3: Solve for \( x \) in terms of \( y \)
To find the inverse, we need to express \( x \) in terms of \( y \). First, square both sides to eliminate the square root:
\[ y^2 = 2x - 3 \]

### Step 4: Rearrange the equation
Now, rearranging the equation to isolate \( x \):
\[ 2x = y^2 + 3 \]
\[ x = \frac{y^2 + 3}{2} \]

### Step 5: Write the inverse function
Since we have expressed \( x \) in terms of \( y \), we can write the inverse function:
\[ f^{-1}(x) = \frac{x^2 + 3}{2} \]

### Step 6: Prove that \( (f \circ f^{-1})(x) = x \)
Now we need to prove that \( f(f^{-1}(x)) = x \). We will substitute \( f^{-1}(x) \) into \( f(x) \):
\[ f(f^{-1}(x)) = f\left(\frac{x^2 + 3}{2}\right) \]

### Step 7: Substitute into the function
Now, substituting into the function \( f \):
\[ f\left(\frac{x^2 + 3}{2}\right) = \sqrt{2 \left(\frac{x^2 + 3}{2}\right) - 3} \]

### Step 8: Simplify the expression
Simplifying the expression inside the square root:
\[ = \sqrt{x^2 + 3 - 3} \]
\[ = \sqrt{x^2} \]
\[ = |x| \]

### Step 9: Determine the domain
Since \( f(x) = \sqrt{2x - 3} \) is defined for \( x \geq \frac{3}{2} \), we can assume \( x \) is non-negative in this context. Therefore, \( |x| = x \).

### Conclusion
Thus, we have shown that:
\[ f(f^{-1}(x)) = x \]
This proves that \( (f \circ f^{-1})(x) = x \).