To write the equation of the tangent line to the function $f(x)$ at the point where $x = 4$, we can use the point-slope form of a line equation:

$y - y_1 = m(x - x_1)$

Where:

• $(x_1, y_1)$ is the point on the function.
• $m$ is the slope of the tangent line, which is given by $f'(x_1)$.

We know the following information from the problem:

• $f(4) = 17$ gives us the point $(4, 17)$.
• $f'(4) = -3$ gives us the slope of the tangent line at $x = 4$.

Using the point-slope form:

$y - 17 = -3(x - 4)$

This simplifies to:

$y = -3(x - 4) + 17$

$y = -3x + 12 + 17$

$y = -3x + 29$

Thus, the equation of the tangent line is:

$y = -3x + 29$

Would you like more details or have any questions about this?

Here are 5 related questions for further practice:

1. How would the tangent line change if $f'(4) = 5$ instead of $-3$?
2. What is the significance of $f''(4) = 2$ in this context?
3. How can we use the second derivative to analyze the concavity at $x = 4$?
4. If $f(x)$ had a higher-order derivative at $x = 4$, how could we approximate $f(x)$?
5. Can you find the equation of the normal line to $f(x)$ at $x = 4$?

Tip: The second derivative $f''(x)$ provides information about the concavity of the function, which helps in understanding the behavior of the graph around the point of tangency.