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?

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!