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The problem asks you to use the definition of the derivative to find \( f'(x) \), where the given function is:

\[
f(x) = \frac{100}{x}
\]

Then, it asks you to evaluate the derivative at specific points: \( f'(-5) \), \( f'(0) \), and \( f'(4) \) where the derivative exists.

### Step 1: Definition of the Derivative
The definition of the derivative is given by the limit:

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

For \( f(x) = \frac{100}{x} \), we substitute this into the limit definition.

### Step 2: Apply the definition

Substitute \( f(x) = \frac{100}{x} \) into the derivative formula:

\[
f'(x) = \lim_{h \to 0} \frac{\frac{100}{x+h} - \frac{100}{x}}{h}
\]

We now simplify the expression inside the limit:

\[
f'(x) = \lim_{h \to 0} \frac{100 \left(\frac{1}{x+h} - \frac{1}{x}\right)}{h}
\]

The term inside the parentheses can be simplified by finding a common denominator:

\[
\frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)}
\]

Thus, we have:

\[
f'(x) = \lim_{h \to 0} \frac{100 \cdot \frac{-h}{x(x+h)}}{h}
\]

Simplifying:

\[
f'(x) = \lim_{h \to 0} \frac{-100h}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{-100}{x(x+h)}
\]

Finally, as \( h \to 0 \), we get:

\[
f'(x) = \frac{-100}{x^2}
\]

### Step 3: Evaluate at the given points

1. \( f'(-5) \):
   \[
   f'(-5) = \frac{-100}{(-5)^2} = \frac{-100}{25} = -4
   \]

2. \( f'(0) \):
   The derivative \( f'(x) = \frac{-100}{x^2} \) does not exist at \( x = 0 \) because division by zero is undefined. Hence, \( f'(0) \) does not exist.

3. \( f'(4) \):
   \[
   f'(4) = \frac{-100}{4^2} = \frac{-100}{16} = -6.25
   \]

### Final Answers:
- \( f'(-5) = -4 \)
- \( f'(0) \) does not exist.
- \( f'(4) = -6.25 \)

Would you like further clarification or details on any of these steps?

Here are some related questions for practice:
1. What is the geometric interpretation of the derivative at \( x = -5 \) and \( x = 4 \)?
2. How can we interpret the fact that \( f'(0) \) does not exist?
3. Can you use the power rule to differentiate \( f(x) = \frac{100}{x} \) instead of the definition?
4. What are the conditions under which a derivative does not exist?
5. How would the derivative change if the function was \( f(x) = \frac{100}{x^2} \)?

**Tip**: Always check the domain of a function before finding its derivative to identify any points where the derivative may not exist.

Using the definition of the derivative, find f'(x). Then find f'(-5), f'(0), and f'(4) when the derivative exists. The given function is f(x) = 100 / x.

Learn how to find the derivative of f(x) = 100/x using the definition of the derivative. Evaluate f'(-5), f'(0), and f'(4) and understand where the derivative exists. This problem involves key calculus concepts such as limits and rational function derivatives. Suitable for students in Grades 11-12.

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