Let's solve **Question 5.1** using the graph of \( p(x) \): ### a. At which values of \( x \) is the derivative of \( p(x) \) equal to zero? **Explanation:** The derivative \( p'(x) \) represents the slope of the tangent line. \( p'(x) = 0 \) where the tangent is horizontal, i.e., at local maxima or minima (turning points). **From the graph:** - The turning points (where the curve flattens out) are at \( x = -2 \), \( x = 1 \), and \( x = 3 \). **Final Answer:** \[ \boxed{x = -2,\, 1,\, 3} \] --- ### b. Determine where \( p'(x) > 0 \). **Explanation:** \( p'(x) > 0 \) where the function \( p(x) \) is **increasing** (going up as \( x \) increases). **From the graph:** - The function increases between \( x = -2 \) and \( x = 1 \) (from the minimum at \( -2 \) to the maximum at \( 1 \)). - It also increases for \( x > 3 \). **Final Answer:** \[ \boxed{-2 < x < 1 \quad \text{and} \quad x > 3} \] --- ### c. Determine where \( p'(x) < 0 \). **Explanation:** \( p'(x) < 0 \) where the function \( p(x) \) is **decreasing** (going down as \( x \) increases). **From the graph:** - The function decreases for \( x < -2 \). - It also decreases between \( x = 1 \) and \( x = 3 \) (from the maximum at 1 to the minimum at 3). **Final Answer:** \[ \boxed{x < -2 \quad \text{and} \quad 1 < x < 3} \] --- **Summary Table:** | Interval | \( p'(x) \) sign | Explanation | |------------------|------------------|-------------------------| | \( x < -2 \) | Negative | Decreasing | | \( -2 < x < 1 \) | Positive | Increasing | | \( 1 < x < 3 \) | Negative | Decreasing | | \( x > 3 \) | Positive | Increasing | Let me know if you want the solutions for the next part!

Certainly! Let's proceed with the next questions step-by-step. --- ## **Question 5.2** *Using the fact that the derivative represents the slope of the tangent line to a graph, answer:* **Calculate the slope of the tangent line to \( f(z) = -z^2 \) at \( z = 4 \).** ### Step 1: Find the derivative \( f'(z) \) Given \( f(z) = -z^2 \), by differentiation: \[ f'(z) = \frac{d}{dz}(-z^2) = -2z \] --- ### Step 2: Evaluate \( f'(z) \) at \( z = 4 \) \[ f'(4) = -2 \times 4 = -8 \] ### **Final answer:** \[ \boxed{-8} \] **Interpretation:** The slope of the tangent line to \( f(z) \) at \( z=4 \) is \(-8\), indicating the function is decreasing rapidly at that point. --- ## **Question 5.3** *Calculate the instantaneous rate of change at \( z = 1 \) of the function \( f(z) = y^z \ln(2z - 1) \).* *(Note: There seems to be a typo or formatting issue in the original question. Assuming the function is \( f(z) = y^z \ln(2z - 1) \), where \( y \) is a constant, and the task is to differentiate with respect to \( z \).)* ### **Step 1: Recognize the function structure** Assuming \( y \) is a constant, \[ f(z) = y^z \ln(2z - 1) \] --- ### **Step 2: Differentiate \( f(z) \) using product rule** Let: \[ u(z) = y^z \quad \text{and} \quad v(z) = \ln(2z - 1) \] Then: \[ f(z) = u(z) \times v(z) \] ### **Step 3: Find derivatives** - \( u(z) = y^z \) \[ u'(z) = y^z \ln y \] - \( v(z) = \ln(2z - 1) \) \[ v'(z) = \frac{2}{2z - 1} \] --- ### **Step 4: Apply product rule** \[ f'(z) = u'(z) v(z) + u(z) v'(z) = y^z \ln y \times \ln(2z - 1) + y^z \times \frac{2}{2z - 1} \] ### **Step 5: Evaluate at \( z=1 \)** Calculate each component: - \( y^1 = y \) - \( \ln y \) (depends on the constant \( y \), but since \( y \) is unspecified, keep as \( \ln y \)) - \( \ln(2 \times 1 - 1) = \ln(2 - 1) = \ln 1 = 0 \) - \( 2 / (2 \times 1 - 1) = 2 / 1 = 2 \) Putting it together: \[ f'(1) = y \ln y \times 0 + y \times 2 = 0 + 2y = 2y \] --- ### **Final answer:** \[ \boxed{2 y} \] *(Without a specific value for \( y \), the derivative at \( z=1 \) is \( 2 y \).)* --- ## **Question 5.4** *Given a continuous and differentiable function \( k \) on \([0, 2\pi]\), with some values provided, answer:* ### **a. Calculate the rate of change of the vertical position at \( t=3 \) seconds.** *(Assuming the position function is \( g(t) = \sin(t) \), based on the pattern in the question. If the actual function is different, please clarify.)* **Since the exact function isn't fully specified,** let's **assume** \( g(t) = \sin(t) \). - At \( t=3 \), the rate of change is: \[ g'(t) = \cos(t) \] - At \( t=3 \): \[ g'(3) = \cos(3) \] Using a calculator: \[ \cos(3) \approx -0.990 \] **Answer:** \[ \boxed{-0.990} \] --- ### **b. Explain what this rate means in the context of the drone’s motion.** **Interpretation:** The rate of change of the vertical position at \( t=3 \) seconds is approximately \(-0.990\) meters per second, meaning **the drone is descending at nearly 1 meter per second** at that moment. --- ### **c. At what time will the drone momentarily stop ascending or descending?** **In the context of the function:** The drone stops ascending or descending when its vertical velocity is zero, i.e., when the derivative of its position function is zero. For \( g(t) = \sin(t) \): \[ g'(t) = \cos(t) = 0 \] - \( \cos(t) = 0 \) at: \[ t = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \] Within the interval \([0, 2\pi]\): - \( t = \frac{\pi}{2} \) (approximately 1.57 seconds) - \( t = \frac{3\pi}{2} \) (approximately 4.71 seconds) **Final answer:** \[ \boxed{t = \frac{\pi}{2} \text{ seconds} \quad \text{and} \quad t = \frac{3\pi}{2} \text{ seconds}} \] --- ## **Question 6.2** *Calculate \( f'(1) \) if \( f(z) = 2482 \).* ### **Observation:** If \( f(z) = 2482 \), it's a **constant function**. - The derivative of a constant function is: \[ f'(z) = 0 \] - At \( z=1 \), the derivative remains: \[ f'(1) = 0 \] --- # **Summary of Final Answers:** 1. **5.1:** \( x = -2, 1, 3 \) 2. **5.2:** \( -8 \) 3. **5.3:** \( 2 y \) (assuming \( y \) is a constant) 4. **5.4a:** approximately **\(-0.990\)** 5. **5.4b:** The drone is descending at about 1 m/sec at \( t=3 \) seconds. 6. **5.4c:** \( t = \frac{\pi}{2} \) and \( t= \frac{3\pi}{2} \) 7. **6.2:** \( 0 \) Let me know if you'd like further clarification on any part!