use the definition of the derivative to find the derivative of the function f(x) = 1/ 2 square 7x. then find all x-values (if any) where the tangent line is horizontal. if the tangent line is horizontal for all x write r for your anser. if the tangent line is never horizontal, write none for your answer.

To find the derivative of the function \( f(x) = \frac{1}{2} \sqrt{7x} \) using the definition of the derivative, we follow the steps:

### 1. **Definition of the Derivative**:
The derivative of a function \( f(x) \) is given by:

\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]

### 2. **Apply the function to the limit definition**:
We apply the function \( f(x) = \frac{1}{2} \sqrt{7x} \) in this definition.

\[
f'(x) = \lim_{h \to 0} \frac{\frac{1}{2} \sqrt{7(x+h)} - \frac{1}{2} \sqrt{7x}}{h}
\]

### 3. **Simplify the expression**:
Factor out \( \frac{1}{2} \) from the numerator:

\[
f'(x) = \frac{1}{2} \lim_{h \to 0} \frac{\sqrt{7(x+h)} - \sqrt{7x}}{h}
\]

We need to simplify the expression involving the square roots. We multiply both the numerator and the denominator by the conjugate of the numerator:

\[
f'(x) = \frac{1}{2} \lim_{h \to 0} \frac{\left( \sqrt{7(x+h)} - \sqrt{7x} \right) \left( \sqrt{7(x+h)} + \sqrt{7x} \right)}{h \left( \sqrt{7(x+h)} + \sqrt{7x} \right)}
\]

This simplifies the numerator to:

\[
\left( \sqrt{7(x+h)} \right)^2 - \left( \sqrt{7x} \right)^2 = 7(x+h) - 7x = 7h
\]

Now, the expression becomes:

\[
f'(x) = \frac{1}{2} \lim_{h \to 0} \frac{7h}{h \left( \sqrt{7(x+h)} + \sqrt{7x} \right)}
\]

Cancel out \( h \) from the numerator and denominator:

\[
f'(x) = \frac{1}{2} \lim_{h \to 0} \frac{7}{\sqrt{7(x+h)} + \sqrt{7x}}
\]

As \( h \to 0 \), \( \sqrt{7(x+h)} \to \sqrt{7x} \), so the limit simplifies to:

\[
f'(x) = \frac{1}{2} \cdot \frac{7}{2\sqrt{7x}} = \frac{7}{4\sqrt{7x}}
\]

Thus, the derivative of the function is:

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

### 4. **Find when the tangent line is horizontal**:
The tangent line is horizontal when the derivative equals 0. So we solve:

\[
f'(x) = \frac{7}{4\sqrt{7x}} = 0
\]

However, this expression can never be 0 because the numerator is 7, a constant. Therefore, the tangent line is never horizontal for any \( x \).

### Final answer:
The tangent line is **never** horizontal, so the answer is:

\[
\boxed{\text{none}}
\]

---

Would you like further details or have any questions?

Here are some follow-up questions for you:
1. How would you interpret the meaning of the derivative geometrically?
2. Can you explain what the limit definition of a derivative represents?
3. What other methods can be used to find the derivative of the function without using the limit definition?
4. How does the behavior of the square root function affect the derivative?
5. What would happen if we had a different power of \( x \) in the original function?

**Tip:** Always check if the derivative can be zero when determining where the tangent line is horizontal!

Use the definition of the derivative to find the derivative of the function f(x) = 1/2 √7x. Then find all x-values (if any) where the tangent line is horizontal. If the tangent line is horizontal for all x, write 'r' for your answer. If the tangent line is never horizontal, write 'none' for your answer.

Learn how to use the limit definition of the derivative to differentiate f(x) = 1/2 √7x. This calculus problem involves applying the definition of the derivative and exploring when the tangent line is horizontal. Suitable for Grades 11-12.

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