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The problem shown in the image asks for the equation of the tangent line to the graph of \( f(x) = \sqrt{x + 12} \) at the point \( (0, 11) \).

### Steps to solve the problem:

1. **Find the derivative of \( f(x) \):**
   The derivative of the function gives us the slope of the tangent line. First, rewrite \( f(x) \) as:
   \[
   f(x) = (x + 12)^{1/2}
   \]
   To differentiate, apply the chain rule:
   \[
   f'(x) = \frac{1}{2}(x + 12)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x + 12}}
   \]
   
2. **Find the slope at \( x = 0 \):**
   To find the slope of the tangent line at the point \( (0, 11) \), plug \( x = 0 \) into \( f'(x) \):
   \[
   f'(0) = \frac{1}{2\sqrt{0 + 12}} = \frac{1}{2\sqrt{12}} = \frac{1}{2 \cdot 2\sqrt{3}} = \frac{1}{4\sqrt{3}}
   \]
   
   Simplify:
   \[
   f'(0) = \frac{\sqrt{3}}{12}
   \]
   Therefore, the slope of the tangent line is \( \frac{\sqrt{3}}{12} \).

3. **Use the point-slope form of the equation of a line:**
   The point-slope form is:
   \[
   y - y_1 = m(x - x_1)
   \]
   where \( (x_1, y_1) = (0, 11) \) is the point of tangency, and \( m = \frac{\sqrt{3}}{12} \) is the slope.
   
   Substituting these values:
   \[
   y - 11 = \frac{\sqrt{3}}{12}(x - 0)
   \]
   Simplifying:
   \[
   y = \frac{\sqrt{3}}{12}x + 11
   \]

### Conclusion:
The equation of the tangent line is:
\[
y = \frac{\sqrt{3}}{12}x + 11
\]

It matches the answer given in the image. Would you like further clarification or more details on any part of this solution?

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Here are 5 related questions for deeper understanding:

1. How do you apply the chain rule when differentiating composite functions?
2. Why is the slope of the tangent line equal to the derivative at a given point?
3. What are other methods to find the equation of a tangent line?
4. How do you interpret the point-slope form of a line equation geometrically?
5. How would the process change if the function was more complex, like \( f(x) = \sqrt{x^2 + 12} \)?

### Tip:
For finding derivatives of square roots, always rewrite the square root as an exponent to make differentiation easier!

Find the equation of the tangent line to the graph of f(x) = √(x + 12) at the point (0, 11).

Learn how to find the equation of the tangent line to the graph of f(x) = √(x + 12) at the point (0, 11). This problem covers key calculus concepts, including derivatives, the chain rule, and the point-slope form of a line. Suitable for high school or college-level students.

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