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The function given is:

\[
f(x) = 8 + 7\sqrt{x}
\]

We are asked to find:

1. The derivative \( f'(x) \)
2. Evaluate \( f'(1) \)
3. Evaluate \( f'(2) \)
4. Evaluate \( f'(3) \)

### 1. Finding the derivative \( f'(x) \):

The derivative of a constant is 0, so the derivative of 8 is 0. Now, for the term \( 7\sqrt{x} \), we rewrite it as \( 7x^{1/2} \), and apply the power rule.

\[
\frac{d}{dx} [x^{n}] = nx^{n-1}
\]

Thus,

\[
f'(x) = 0 + 7 \cdot \frac{1}{2} x^{-\frac{1}{2}} = \frac{7}{2} \cdot \frac{1}{\sqrt{x}} = \frac{7}{2\sqrt{x}}
\]

### 2. Evaluate \( f'(1) \):

Substitute \( x = 1 \) into the derivative:

\[
f'(1) = \frac{7}{2\sqrt{1}} = \frac{7}{2}
\]

### 3. Evaluate \( f'(2) \):

Substitute \( x = 2 \) into the derivative:

\[
f'(2) = \frac{7}{2\sqrt{2}} = \frac{7}{2\sqrt{2}} = \frac{7\sqrt{2}}{4}
\]

### 4. Evaluate \( f'(3) \):

Substitute \( x = 3 \) into the derivative:

\[
f'(3) = \frac{7}{2\sqrt{3}} = \frac{7\sqrt{3}}{6}
\]

### Final answers:

- \( f'(x) = \frac{7}{2\sqrt{x}} \)
- \( f'(1) = \frac{7}{2} \)
- \( f'(2) = \frac{7\sqrt{2}}{4} \)
- \( f'(3) = \frac{7\sqrt{3}}{6} \)

Would you like more details on any part of this process?

Here are five related questions you might want to explore:

1. How do we apply the power rule for different powers of \( x \)?
2. What is the chain rule, and when should we use it in derivatives?
3. Can we use this method for more complicated radicals, such as \( \sqrt[3]{x} \)?
4. What is the general derivative of \( f(x) = a + b\sqrt{x} \)?
5. How would the derivative change if the radical term were in the denominator?

**Tip:** Always simplify radical expressions after differentiating. This can help with both interpretation and further calculations.

Find the derivative of f(x) = 8 + 7√x, then evaluate f'(1), f'(2), and f'(3).

Learn how to differentiate the function f(x) = 8 + 7√x and evaluate the derivative at specific points: f'(1), f'(2), and f'(3). This exercise covers key differentiation concepts, including the power rule for radicals. Suitable for high school students in grades 10-12.

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