Let $$ f(x)=4+\sqrt{6 x-7} $$. Find $$ f^{-1}(x) $$ $$ f^{-1}(x)=\square $$, for $$ x \geq 4 $$

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Let $$ y = f(x) = 4 + \sqrt{6x - 7} $$.

1. Isolate the square root term:
   $$
   y - 4 = \sqrt{6x - 7}
   $$

2. Square both sides to remove the square root:
   $$
   (y - 4)^2 = 6x - 7
   $$

3. Solve for $$ x $$:
   $$
   6x = (y - 4)^2 + 7 \quad \Longrightarrow \quad x = \frac{(y-4)^2 + 7}{6}
   $$

4. Interchange $$ x $$ and $$ y $$ to find the inverse function:
   $$
   f^{-1}(x) = \frac{(x-4)^2 + 7}{6}
   $$

Thus, the inverse function is:
$$
f^{-1}(x) = \frac{(x-4)^2 + 7}{6}, \quad \text{for } x \geq 4.
$$
$$ f^{-1}(x) = \frac{(x-4)^2 + 7}{6} $$, for $$ x \geq 4 $$.

Let \( f(x)=4+\sqrt{6 x-7} \). Find \( f^{-1}(x) \) \( f^{-1}(x)=\square \), for \( x \geq 4 \)

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