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The image asks to find the equation of the tangent line to the graph of \( f(x) = \sqrt{2x + 1} \) at \( x = 3 \).

### Solution:

The equation of the tangent line at a given point \( x = a \) can be written as:
\[
y - f(a) = f'(a)(x - a)
\]
where \( f'(a) \) is the derivative of the function at \( x = a \).

#### Step 1: Compute the derivative \( f'(x) \)

Given \( f(x) = \sqrt{2x + 1} \), let's apply the chain rule to find the derivative.

\[
f'(x) = \frac{d}{dx} \left( (2x + 1)^{1/2} \right) = \frac{1}{2}(2x + 1)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x + 1}}
\]

So the derivative is:
\[
f'(x) = \frac{1}{\sqrt{2x + 1}}
\]

#### Step 2: Evaluate \( f(3) \) and \( f'(3) \)

Now, evaluate the function and its derivative at \( x = 3 \):
\[
f(3) = \sqrt{2(3) + 1} = \sqrt{7}
\]
\[
f'(3) = \frac{1}{\sqrt{2(3) + 1}} = \frac{1}{\sqrt{7}}
\]

#### Step 3: Write the equation of the tangent line

Using the point \( (3, \sqrt{7}) \) and the slope \( \frac{1}{\sqrt{7}} \), the equation of the tangent line is:

\[
y - \sqrt{7} = \frac{1}{\sqrt{7}}(x - 3)
\]

Simplifying:

\[
y = \frac{1}{\sqrt{7}}(x - 3) + \sqrt{7}
\]

This is the equation of the tangent line at \( x = 3 \).

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Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How does the chain rule apply to more complex functions?
2. Can the slope of the tangent line be negative depending on the function?
3. What does the tangent line represent geometrically?
4. How would the tangent line equation change for \( f(x) = \sqrt{3x + 2} \)?
5. What happens to the tangent line as \( x \to \infty \) for this function?

**Tip:** Always check the domain of the function before computing the derivative, especially for functions involving square roots.

Determine the equation of the tangent line to the graph of f(x) = √(2x + 1) at x = 3.

Learn how to find the equation of the tangent line to the function f(x) = √(2x + 1) at x = 3. This problem involves calculus concepts such as derivatives and the chain rule. Suitable for students in Grades 11-12 studying calculus.

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