The function provided is:

$f(x) = -2x^2 + 13x$

We are tasked with finding the derivative $f'(x)$, and then calculating specific values of this derivative at $x = 4$, $x = 17$, and $x = -3$.

Step 1: Find the derivative of $f(x)$

The derivative of $f(x)$ with respect to $x$ can be calculated using basic differentiation rules. For $f(x) = -2x^2 + 13x$:

$f'(x) = \frac{d}{dx} (-2x^2 + 13x)$

Using the power rule:

• The derivative of $-2x^2$ is $-4x$,
• The derivative of $13x$ is $13$.

So, the derivative is:

$f'(x) = -4x + 13$

Step 2: Compute the specific values

1. $f'(4)$: $f'(4) = -4(4) + 13 = -16 + 13 = -3$

2. $f'(17)$: $f'(17) = -4(17) + 13 = -68 + 13 = -55$

3. $f'(-3)$: $f'(-3) = -4(-3) + 13 = 12 + 13 = 25$

Conclusion:

• $f'(4) = -3$
• $f'(17) = -55$
• $f'(-3) = 25$

Let me know if you need further clarifications! Here are five questions to deepen your understanding:

1. How do you apply the power rule when differentiating functions?
2. Why does the derivative of a quadratic function result in a linear expression?
3. What is the significance of calculating the derivative at specific points?
4. How does the slope of the function $f(x) = -2x^2 + 13x$ change as $x$ increases?
5. Can you explain why the derivative of a constant is zero?

Tip: Always check your differentiation by revisiting basic rules like the power rule and chain rule to ensure your results are accurate.