Math Problem Statement
Given the function f(x) = {3 - x^2 if x ≤ 1, 2/x if x > 1} on [0,2], verify that the hypothesis of the mean-value theorem is satisfied and find a suitable choice for c that satisfies the conclusion of the mean-value theorem.
Solution
Let's analyze the problem step by step.
Given function:
\begin{cases} 3 - x^2 & \text{if } x \leq 1, \\ \frac{2}{x} & \text{if } x > 1. \end{cases}$$ on the interval $$[0, 2]$$. ### 1. Verifying the hypothesis of the Mean Value Theorem (MVT): The Mean Value Theorem requires two conditions: - **Continuity** on the closed interval $$[a, b]$$, - **Differentiability** on the open interval $$(a, b)$$. **Continuity:** - $$f(x) = 3 - x^2$$ is a polynomial, and polynomials are continuous. Therefore, $$f(x)$$ is continuous on $$[0, 1]$$. - $$f(x) = \frac{2}{x}$$ is continuous for $$x > 1$$, specifically on the interval $$(1, 2]$$. - We need to check if the function is continuous at $$x = 1$$, where the two pieces of the function meet. We do this by checking the left-hand and right-hand limits at $$x = 1$$. $$\lim_{x \to 1^-} f(x) = 3 - 1^2 = 2, \quad \lim_{x \to 1^+} f(x) = \frac{2}{1} = 2.$$ Since both limits are equal and match the function's value at $$x = 1$$, the function is continuous at $$x = 1$$. Therefore, $$f(x)$$ is continuous on $$[0, 2]$$. **Differentiability:** - $$f(x) = 3 - x^2$$ is differentiable for $$x \in (0, 1)$$. - $$f(x) = \frac{2}{x}$$ is differentiable for $$x \in (1, 2)$$. - We need to check if $$f(x)$$ is differentiable at $$x = 1$$. To do this, we check if the left-hand and right-hand derivatives at $$x = 1$$ are equal. For $$x \leq 1$$: $$f'(x) = \frac{d}{dx}(3 - x^2) = -2x, \quad f'(1) = -2(1) = -2.$$ For $$x > 1$$: $$f'(x) = \frac{d}{dx}\left(\frac{2}{x}\right) = -\frac{2}{x^2}, \quad f'(1) = -\frac{2}{1^2} = -2.$$ Since the left-hand and right-hand derivatives are equal at $$x = 1$$, $$f(x)$$ is differentiable at $$x = 1$$. Thus, $$f(x)$$ is differentiable on $$(0, 2)$$. **Conclusion:** Both conditions of the Mean Value Theorem are satisfied. --- ### 2. Finding a suitable $$c$$ that satisfies the conclusion of the Mean Value Theorem: The Mean Value Theorem states that there exists $$c \in (0, 2)$$ such that $$f'(c) = \frac{f(2) - f(0)}{2 - 0}.$$ Let's calculate the right-hand side: - $$f(0) = 3 - 0^2 = 3$$, - $$f(2) = \frac{2}{2} = 1$$. Thus, $$\frac{f(2) - f(0)}{2 - 0} = \frac{1 - 3}{2} = \frac{-2}{2} = -1.$$ Now, we need to find $$c \in (0, 2)$$ such that $$f'(c) = -1$$. **For $$x \leq 1$$:** $$f'(x) = -2x.$$ We set $$-2x = -1$$, which gives $$x = \frac{1}{2}$$. Thus, $$c = \frac{1}{2}$$ is a solution in $$(0, 1)$$. **For $$x > 1$$:** $$f'(x) = -\frac{2}{x^2}.$$ We set $$-\frac{2}{x^2} = -1$$, which gives $$x^2 = 2$$, or $$x = \sqrt{2} \approx 1.414$$. Thus, $$c = \sqrt{2}$$ is a solution in $$(1, 2)$$. ### Conclusion: The values $$c = \frac{1}{2}$$ and $$c = \sqrt{2}$$ both satisfy the conclusion of the Mean Value Theorem. --- Would you like more details on any of these steps? Here are some related questions to expand your understanding: 1. How can we visually interpret the Mean Value Theorem? 2. What happens if a function is not continuous or differentiable at a point on the interval? 3. How does the Mean Value Theorem relate to Rolle's Theorem? 4. Can there be more than one value of $$c$$ that satisfies the MVT for a function? 5. What are some applications of the Mean Value Theorem in physics or other fields? **Tip:** Always check both continuity and differentiability carefully, especially when dealing with piecewise functions, as these are common points where these properties may fail.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Mean Value Theorem
Piecewise Functions
Continuity
Differentiability
Formulas
f'(c) = (f(b) - f(a)) / (b - a)
f'(x) for x ≤ 1: -2x
f'(x) for x > 1: -2/x^2
Theorems
Mean Value Theorem
Suitable Grade Level
Undergraduate Level (First-Year Calculus)
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