Let's go through each problem step-by-step: --- # 1. Estimating the value of \( c \) from the graph ### **Given:** - The graph of the function (not explicitly given in the prompt, but based on the instructions, it appears to be a polynomial curve). - The goal: Estimate \( c \) such that the mean value theorem applies on \([-3, 2]\). ### **Step 1:** Find the average value of \( y \) over \([-3, 2]\). The Mean Value Theorem for integrals states: \[ f(c) = \frac{1}{b-a} \int_a^b f(x) \, dx \] where \( a = -3 \) and \( b = 2 \). ### **Step 2:** Approximate the integral \( \int_{-3}^{2} f(x) \, dx \). Since the graph isn't explicitly provided with exact coordinates, **we estimate** the average height of the curve over the interval: - At \( x = -3 \): \( y \approx -4 \) - At \( x = \): \( y \approx 2 \) - At \( x = 2 \): \( y \approx 4 \) Estimate the average value of the function over \([-3, 2]\): \[ \text{Average } y \approx \frac{(-4) + 2 + 4}{3} = \frac{2}{3} \approx .67 \] ### **Step 3:** Compute the average value: \[ f(c) \approx .67 \] ### **Step 4:** Find \( c \) such that \( f(c) \approx .67 \). From the graph, the function crosses the approximate value \( y \approx .67 \) at around: - \( c \approx -1 \) **Therefore, the estimated value of \( c \) is approximately \(-1\).** --- # 2. Find the number \( c \) satisfying the Mean Value Theorem for the function: \[ f(x) = x^3 - 3x^2 - 18x + 40 \] over \([-3, 2]\). ### **Step 1:** Calculate \( f(-3) \) and \( f(2) \). \[ f(-3) = (-3)^3 - 3(-3)^2 - 18(-3) + 40 = -27 - 3(9) + 54 + 40 = -27 - 27 + 54 + 40 = 40 \] \[ f(2) = (2)^3 - 3(2)^2 - 18(2) + 40 = 8 - 3(4) - 36 + 40 = 8 - 12 - 36 + 40 = \] ### **Step 2:** Compute the average rate of change: \[ \frac{f(2) - f(-3)}{2 - (-3)} = \frac{ - 40}{5} = -8 \] ### **Step 3:** Find \( c \in (-3, 2) \) such that \( f'(c) = -8 \). ### **Step 4:** Derive \( f'(x) \): \[ f'(x) = 3x^2 - 6x - 18 \] Set \( f'(c) = -8 \): \[ 3c^2 - 6c - 18 = -8 \] \[ 3c^2 - 6c - 10 = \] ### **Step 5:** Solve the quadratic: \[ 3c^2 - 6c - 10 = \] Using the quadratic formula: \[ c = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 3 \times (-10)}}{2 \times 3} \] \[ c = \frac{6 \pm \sqrt{36 + 120}}{6} = \frac{6 \pm \sqrt{156}}{6} \] \[ c = \frac{6 \pm 2\sqrt{39}}{6} = 1 \pm \frac{\sqrt{39}}{3} \] Approximate \( \sqrt{39} \approx 6.24 \): \[ c \approx 1 \pm \frac{6.24}{3} = 1 \pm 2.08 \] - \( c \approx 1 + 2.08 = 3.08 \) (outside \([-3, 2]\)) - \( c \approx 1 - 2.08 = -1.08 \) (within \([-3, 2]\)) **Final answer:** \[ \boxed{ c \approx -1.08 } \] --- # 3. Show that the equation \( x^3 - 15x + c = \) has at most one zero in \([-2, 2]\). ### **Step 1:** Consider the function: \[ g(x) = x^3 - 15x + c \] ### **Step 2:** Find the derivative: \[ g'(x) = 3x^2 - 15 \] ### **Step 3:** Analyze \( g'(x) \): - \( g'(x) = 3x^2 - 15 \) - \( g'(x) = \Rightarrow x^2 = 5 \Rightarrow x = \pm \sqrt{5} \approx \pm 2.24 \) ### **Step 4:** Since the interval is \([-2, 2]\): - \( g'(x) \) on \([-2, 2]\): At \( x = \pm 2 \): \[ g'(\pm 2) = 3(4) - 15 = 12 - 15 = -3 \] - \( g'(x) \) in \([-2, 2]\), because \( \pm \sqrt{5} \) are outside the interval, \( g'(x) \) does not change sign within \([-2, 2]\). - For \( x \in [-2, 2] \), \( g'(x) < \), so \( g(x) \) is **strictly decreasing** on \([-2, 2]\). ### **Step 5:** Conclusion: Since \( g(x) \) is strictly decreasing on \([-2, 2]\), **it can have at most one zero** in this interval. --- # **Summary of Final Answers** | Problem | Final Result | |---|---| | 1 | \( c \approx -1 \) (estimated from graph) | | 2 | \( c \approx -1.08 \) | | 3 | The equation has **at most one zero** in \([-2, 2]\) | --- Let me know if you'd like further clarification!