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To find the equation of the tangent line to the graph of the function \( f(x) = \frac{\sqrt{x}}{3x + 3} \) at the point \( (3, f(3)) \), we need to follow these steps:

1. **Calculate \( f(3) \)**:  
   Plug \( x = 3 \) into the function to find the y-coordinate of the point of tangency.
   
   \[
   f(3) = \frac{\sqrt{3}}{3(3) + 3} = \frac{\sqrt{3}}{9 + 3} = \frac{\sqrt{3}}{12}
   \]
   
   So, the point of tangency is \( \left(3, \frac{\sqrt{3}}{12}\right) \).

2. **Find the derivative \( f'(x) \)**:  
   The derivative of \( f(x) \) gives the slope of the tangent line. To find \( f'(x) \), we use the quotient rule:
   
   \[
   f(x) = \frac{\sqrt{x}}{3x + 3}
   \]
   
   The quotient rule states:
   \[
   f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}
   \]
   where \( g(x) = \sqrt{x} \) and \( h(x) = 3x + 3 \).

   First, find the derivatives:
   \[
   g'(x) = \frac{1}{2\sqrt{x}}, \quad h'(x) = 3
   \]
   
   Then, apply the quotient rule:
   \[
   f'(x) = \frac{\left(\frac{1}{2\sqrt{x}}\right)(3x + 3) - \left(\sqrt{x}\right)(3)}{(3x + 3)^2}
   \]
   
   Simplify this expression:
   \[
   f'(x) = \frac{\frac{3x + 3}{2\sqrt{x}} - 3\sqrt{x}}{(3x + 3)^2}
   \]
   
   Simplifying further:
   \[
   f'(x) = \frac{\frac{3(x + 1) - 6x}{2\sqrt{x}}}{(3x + 3)^2} = \frac{\frac{3 - 3x}{2\sqrt{x}}}{(3x + 3)^2}
   \]

3. **Evaluate \( f'(3) \)**:  
   Substitute \( x = 3 \) into the derivative to find the slope at \( x = 3 \):
   
   \[
   f'(3) = \frac{\frac{3(1 - 3)}{2\sqrt{3}}}{(3(3) + 3)^2} = \frac{\frac{3(-2)}{2\sqrt{3}}}{12^2} = \frac{-3\sqrt{3}}{72} = \frac{-\sqrt{3}}{24}
   \]
   
   So, the slope of the tangent line at \( x = 3 \) is \( \frac{-\sqrt{3}}{24} \).

4. **Use the point-slope form of the line**:
   The equation of the tangent line is given by:
   
   \[
   y - y_1 = m(x - x_1)
   \]
   
   Here, \( (x_1, y_1) = \left(3, \frac{\sqrt{3}}{12}\right) \) and \( m = \frac{-\sqrt{3}}{24} \).
   
   Substituting these values:
   \[
   y - \frac{\sqrt{3}}{12} = \frac{-\sqrt{3}}{24}(x - 3)
   \]
   
   Simplifying:
   \[
   y = \frac{-\sqrt{3}}{24}x + \frac{\sqrt{3}}{8} + \frac{\sqrt{3}}{12}
   \]
   
   Combine the constant terms:
   \[
   y = \frac{-\sqrt{3}}{24}x + \frac{\sqrt{3}}{24}
   \]

So, the equation of the tangent line is:

\[
y = \frac{-\sqrt{3}}{24}(x - 1)
\]

Would you like further details or have any questions?

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**Related Questions:**

1. How does the quotient rule apply to different types of functions?
2. What is the geometric interpretation of a derivative?
3. How can the slope of a tangent line help in understanding the behavior of a function?
4. What is the importance of finding the tangent line in real-world applications?
5. How would you find the tangent line for more complex functions?

**Tip:** When using the quotient rule, carefully keep track of each component function's derivative to avoid errors.

Find an equation for the line tangent to the graph of f(x) = √x / (3x + 3) at the point (3, f(3)).

This problem demonstrates how to find the equation of the tangent line to the function f(x) = √x / (3x + 3) at the point (3, f(3)). It covers essential calculus concepts, including differentiation using the quotient rule and applying the point-slope form. Suitable for students in Grades 11-12.

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