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From the problem image you uploaded, it appears you are trying to find the equation of the tangent line to the function \( y = f(x) = \sqrt{x} \) at \( x = 18 \).

To determine the equation of the tangent line, you need to find both the slope at the point and the y-coordinate of the function at that point. The slope of the tangent line is given by the derivative of the function evaluated at that point.

### Step 1: Find the derivative
For the function \( f(x) = \sqrt{x} \), its derivative \( f'(x) \) is given by:
\[ f'(x) = \frac{1}{2\sqrt{x}} \]

### Step 2: Evaluate the derivative at \( x = 18 \)
\[ f'(18) = \frac{1}{2\sqrt{18}} = \frac{1}{6\sqrt{2}} \]
(As shown in your uploaded document).

### Step 3: Find the y-coordinate at \( x = 18 \)
\[ f(18) = \sqrt{18} = 3\sqrt{2} \]

### Step 4: Use the point-slope form of a line
The point-slope form of a line is given by:
\[ y - y_1 = m(x - x_1) \]
Where \( m \) is the slope, and \( (x_1, y_1) \) is the point through which the line passes. Using the point \( (18, 3\sqrt{2}) \) and slope \( \frac{1}{6\sqrt{2}} \):
\[ y - 3\sqrt{2} = \frac{1}{6\sqrt{2}}(x - 18) \]

### Step 5: Simplify the equation
\[ y = \frac{1}{6\sqrt{2}}x - \frac{18}{6\sqrt{2}} + 3\sqrt{2} \]
\[ y = \frac{1}{6\sqrt{2}}x - \frac{3\sqrt{2}}{2} + 3\sqrt{2} \]
\[ y = \frac{1}{6\sqrt{2}}x + \frac{3\sqrt{2}}{2} \]

This is the equation of the tangent line to the curve \( y = \sqrt{x} \) at \( x = 18 \).

Do you need any more details or have other questions on this? Here are some related questions you might consider:

1. How would the tangent line change if evaluated at a different point?
2. What is the significance of the tangent line in understanding the behavior of the function?
3. Could you determine the tangent line using a numerical method?
4. How does the derivative change as \( x \) increases?
5. What are other methods for finding the slope of a tangent line without using derivatives?

**Tip:** When solving for the equation of a tangent line, it's helpful to first verify the function's behavior around the point of tangency to ensure the function is differentiable at that point.

Determine the equation of the line tangent to y = f(x) = √x at x = 18.

Explore how to find the equation of the tangent line to the curve y = sqrt(x) at x = 18 using calculus concepts such as derivatives and limits. This page offers a detailed solution applying the point-slope form and the derivative of sqrt(x).

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