To provide accurate answers to these questions, we'll need to assume or be given a specific function \( f(x) \). However, I will illustrate the process for solving each part with a general function \( f(x) \).
### Part A: Average Rate of Change and Instantaneous Rate of Change
1. **Average Rate of Change of \( f \) over \([-8, 1]\)**
The average rate of change of \( f \) over the interval \([-8, 1]\) is given by the formula:
\[
{Average Rate of Change} = \frac{f(1) - f(-8)}{1 - (-8)} = \frac{f(1) - f(-8)}{9}
\]
2. **Instantaneous Rate of Change**
The instantaneous rate of change is given by the derivative \( f'(x) \). According to the Mean Value Theorem (MVT), if \( f \) is continuous on \([-8, 1]\) and differentiable on \((-8, 1)\), then there exists at least one \( c \in (-8, 1) \) such that:
\[
f'(c) = \frac{f(1) - f(-8)}{1 - (-8)} = \frac{f(1) - f(-8)}{9}
\]
This tells us that the instantaneous rate of change \( f'(c) \) equals the average rate of change at least once in the interval. The exact number of times they are equal depends on the function \( f(x) \). For a typical function, they are equal exactly once, but for more complex functions, this can vary.
### Part B: Equation of the Tangent Line at \( x = -3.5 \)
To find the equation of the tangent line at \( x = -3.5 \):
1. Find the derivative \( f'(x) \).
2. Evaluate the derivative at \( x = -3.5 \) to find the slope \( m \).
3. The point of tangency is \((-3.5, f(-3.5))\).
The equation of the tangent line is:
\[
y - f(-3.5) = f'(-3.5)(x + 3.5)
\]
This simplifies to:
\[
y = f'(-3.5)(x + 3.5) + f(-3.5)
\]
### Part C: Finding Limits
1. **Limit as \( x \) approaches \(-1\)**
\[
\lim_{x \to -1} \frac{f(x) - f(-1)}{x - (-1)} = f'(-1)
\]
This limit represents the derivative of \( f \) at \( x = -1 \).
2. **Limit as \( x \) approaches \(-5\)**
\[
\lim_{x \to -5} \frac{f(x) - f(-5)}{x - (-5)} = f'(-5)
\]
This limit represents the derivative of \( f \) at \( x = -5 \).
Both limits exist if the derivatives \( f'(-1) \) and \( f'(-5) \) exist. If \( f \) is differentiable at \( x = -1 \) and \( x = -5 \), these limits exist and are equal to the values of the derivatives at those points.
### Summary
- **Part A**: Calculate \(\frac{f(1) - f(-8)}{9}\). By the Mean Value Theorem, \( f'(c) \) will be equal to this average rate of change at least once in \((-8, 1)\).
- **Part B**: The tangent line at \( x = -3.5 \) has the equation \( y = f'(-3.5)(x + 3.5) + f(-3.5) \).
- **Part C**: The limits are \( f'(-1) \) and \( f'(-5) \), provided \( f \) is differentiable at those points.
