Certainly! Let's through each part step-by-step. --- # Part 1: Estimating \(c\) using the graph ### Task: Estimate the value of \(c\) that satisfies the conclusion of the Mean Value Theorem (MVT) on the interval \([-3, 2]\). ### Explanation: - The MVT states that if \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists some \(c \in (a, b)\) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] - To estimate \(c\), we need: - \(f(a)\) at \(a = -3\) - \(f(b)\) at \(b = 2\) - The average rate of change over \([-3, 2]\) --- ### Step 1: Find \(f(-3)\) and \(f(2)\) from the graph - **At \(x = -3\)**: - From the graph, \(f(-3) \approx \) - **At \(x = 2\)**: - From the graph, \(f(2) \approx -2\) ### Step 2: Compute the average rate of change \[ \frac{f(2) - f(-3)}{2 - (-3)} = \frac{-2 - }{2 + 3} = \frac{-2}{5} = -.4 \] ### Step 3: Find \(c\) where the slope of the tangent equals \(-.4\) - On the graph, look for a point \(c \in (-3, 2)\) where the tangent line has a slope of approximately \(-.4\). - **Estimate from the graph**: - Around \(x \approx \), the slope looks close to \(-.4\). - Alternatively, at \(x \approx 1\), the slope appears flatter, so around \(\) to \(1\), the slope is about \(-.4\). --- ### **Answer for Part 1:** **Estimated \(c \approx \) to \(1\)** --- # Part 2: Find \(c\) such that the Mean Value Theorem holds for \(f(x) = x^3 - 3x^2 - 18x + 40\) ### Interval: \([-3, 2]\) ### Step 1: Compute \(f(-3)\) and \(f(2)\) - \(f(x) = x^3 - 3x^2 - 18x + 40\) - **At \(x = -3\)**: \[ f(-3) = (-3)^3 - 3(-3)^2 - 18(-3) + 40 = -27 - 3(9) + 54 + 40 = -27 - 27 + 54 + 40 = 40 \] - **At \(x = 2\)**: \[ f(2) = 8 - 3(4) - 36 + 40 = 8 - 12 - 36 + 40 = \] ### Step 2: Calculate the average rate of change \[ \frac{f(2) - f(-3)}{2 - (-3)} = \frac{ - 40}{5} = \frac{-40}{5} = -8 \] ### Step 3: Find \(c\) such that \(f'(c) = -8\) - First, find \(f'(x)\): \[ f'(x) = 3x^2 - 6x - 18 \] - Set \(f'(c) = -8\): \[ 3c^2 - 6c - 18 = -8 \] \[ 3c^2 - 6c - 18 + 8 = \] \[ 3c^2 - 6c - 10 = \] - Divide through by 3: \[ c^2 - 2c - \frac{10}{3} = \] - Use quadratic formula: \[ c = \frac{2 \pm \sqrt{(2)^2 - 4 \times 1 \times (-10/3)}}{2} \] \[ c = \frac{2 \pm \sqrt{4 + \frac{40}{3}}}{2} = \frac{2 \pm \sqrt{\frac{12 + 40}{3}}}{2} = \frac{2 \pm \sqrt{\frac{52}{3}}}{2} \] \[ c = \frac{2 \pm \sqrt{\frac{156}{9}}}{2} = \frac{2 \pm \frac{\sqrt{156}}{3}}{2} \] - Simplify \(\sqrt{156}\): \[ \sqrt{156} \approx 12.49 \] - Now, \[ c = \frac{2 \pm \frac{12.49}{3}}{2} = \frac{2 \pm 4.16}{2} \] - **Solutions**: \[ c_1 = \frac{2 + 4.16}{2} = \frac{6.16}{2} = 3.08 \] \[ c_2 = \frac{2 - 4.16}{2} = \frac{-2.16}{2} = -1.08 \] ### Step 4: Check if \(c\) is in \([-3, 2]\): - \(c_1 \approx 3.08\) → outside interval - \(c_2 \approx -1.08\) → inside interval --- ### **Final answer for Part 2:** **\(c \approx -1.08\)** --- # Part 3: Show that \(x^3 - 15x + c = \) has at most one zero in \([-2, 2]\) ### Step 1: Find the derivative to analyze the function's monotonicity \[ f(x) = x^3 - 15x + c \] \[ f'(x) = 3x^2 - 15 \] ### Step 2: Find critical points: \[ f'(x) = \Rightarrow 3x^2 - 15 = \Rightarrow x^2 = 5 \Rightarrow x = \pm \sqrt{5} \approx \pm 2.236 \] ### Step 3: Check whether critical points are in \([-2, 2]\) - \(\sqrt{5} \approx 2.236\), which is **outside** \([-2, 2]\). - \(-\sqrt{5} \approx -2.236\), **outside** \([-2, 2]\). ### Step 4: Monotonicity in \([-2, 2]\) - Since \(f'(x) = 3x^2 - 15\): - For \(x \in [-2, 2]\): \[ x^2 \leq 4 \] \[ 3x^2 \leq 12 \] \[ 3x^2 - 15 \leq 12 - 15 = -3 \] - Therefore, \(f'(x) < \) for all \(x \in [-2, 2]\). - **Conclusion**: \(f(x)\) is strictly decreasing on \([-2, 2]\). ### **Implication:** - Since \(f(x)\) is strictly decreasing on \([-2, 2]\), it can have **at most one zero** in this interval. --- # **Summary of Final Answers:** 1. **Estimated \(c\) for the graph:** approximately ** to 1**. 2. **Value of \(c\) satisfying MVT for \(f(x) = x^3 - 3x^2 - 18x + 40\):** approximately **\(-1.08\)**. 3. **Number of zeros of \(x^3 - 15x + c=\) in \([-2, 2]\):** **at most one**. --- Let me know if you'd like further clarification!