Let's analyze the problem and solve for both $f(5)$ and $f'(5)$.

Step 1: Understanding the problem

We are given:

• The function $y = f(x)$, where $f(5) = 3$.
• The tangent line to $y = f(x)$ at the point $(5, 3)$ passes through another point $(0, 2)$.

We are tasked with finding:

• $f(5)$, which is given directly as $f(5) = 3$.
• $f'(5)$, which is the slope of the tangent line at $x = 5$.

Step 2: Finding the slope $f'(5)$

The slope of the tangent line can be calculated using the formula for the slope between two points. The tangent line passes through the points $(5, 3)$ and $(0, 2)$. The slope $m = f'(5)$ is given by: $f'(5) = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 3}{0 - 5} = \frac{-1}{-5} = \frac{1}{5}$

So, $f'(5) = \frac{1}{5}$.

Step 3: Final answer

• $f(5) = 3$
• $f'(5) = \frac{1}{5}$

Would you like any more details or have further questions?

Here are 5 related questions to expand this topic:

1. How do we find the equation of the tangent line using the point-slope form?
2. What is the geometric meaning of the derivative at a point?
3. How would the tangent line equation change if $f'(5)$ was negative?
4. What is the difference between finding a secant line and a tangent line slope?
5. How would the process differ if you were given different points through which the tangent passes?

Tip: Remember that the slope of the tangent line at a point gives you the rate of change of the function at that point.