10) If $$ f(x)=\frac{1}{x-a}+b $$, then $$ f^{-1}(a)+f(b)= $$

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To solve the problem, let's find the inverse function $$ f^{-1}(y) $$ and then compute $$ f^{-1}(a) + f(b) $$.

Given:
$$ f(x) = \frac{1}{x - a} + b $$

1. **Find the inverse function $$ f^{-1}(y) $$:**

Start by setting $$ y = f(x) $$:
   $$
   y = \frac{1}{x - a} + b
   $$
   
   Solve for $$ x $$:
   $$
   y - b = \frac{1}{x - a}
   $$
   $$
   x - a = \frac{1}{y - b}
   $$
   $$
   x = a + \frac{1}{y - b}
   $$
   
   Therefore, the inverse function is:
   $$
   f^{-1}(y) = a + \frac{1}{y - b}
   $$

2. **Compute $$ f^{-1}(a) $$:**
   $$
   f^{-1}(a) = a + \frac{1}{a - b}
   $$

3. **Compute $$ f(b) $$:**
   $$
   f(b) = \frac{1}{b - a} + b = -\frac{1}{a - b} + b
   $$

4. **Add $$ f^{-1}(a) $$ and $$ f(b) $$:**
   $$
   f^{-1}(a) + f(b) = \left(a + \frac{1}{a - b}\right) + \left(-\frac{1}{a - b} + b\right) = a + b
   $$

**Final Answer:**
$$
a + b
$$
$$ f^{-1}(a) + f(b) = a + b $$

If , then

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