Problem Breakdown:

We are tasked with finding the derivative $f'(5)$, which represents the slope of the tangent line to the function $f$ at the point $(5, -7)$. The tangent line also passes through the point $(1, -1)$.

Solution:

1. Slope of the Line: The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is calculated as: $m = \frac{y_2 - y_1}{x_2 - x_1}.$

   Substituting the given points $(5, -7)$ and $(1, -1)$: $m = \frac{-1 - (-7)}{1 - 5} = \frac{-1 + 7}{1 - 5} = \frac{6}{-4} = -\frac{3}{2}.$

   So, the slope of the tangent line, and thus $f'(5)$, is: $f'(5) = -\frac{3}{2}.$

Final Answer:

$f'(5) = -\frac{3}{2}.$

Would you like me to explain any step in more detail?

Related Questions:

1. How do you calculate the equation of the tangent line using the point-slope form?
2. Why is the derivative $f'(x)$ at a point equal to the slope of the tangent line?
3. What happens if the tangent line passes through other points? How do you handle such cases?
4. Can a function have more than one tangent line at a given point?
5. How do you extend this concept to find the tangent line for parametric or implicit functions?

Tip:

When working with tangent line problems, always identify the key points and compute the slope carefully to avoid sign errors.