Math Problem Statement
d:10,3]→R defined by /(x)=(0, x < 1 x >= 1 lmvt is applicable
Solution
To determine whether the Lagrange Mean Value Theorem (LMVT) is applicable to the function defined by
\begin{cases} 0, & x < 1 \\ x, & x \geq 1 \end{cases}$$ let’s review the conditions under which LMVT can be applied. ### Conditions for the Lagrange Mean Value Theorem (LMVT): 1. The function must be **continuous** on the closed interval $$[a, b]$$. 2. The function must be **differentiable** on the open interval $$(a, b)$$. Given that LMVT applies to intervals of the form $$[a, b]$$ where $$a < b$$, it seems there might be an issue with the given interval $$(10, 3]$$ as $$10 > 3$$. This notation might be an error because it should typically be in the increasing order, such as $$[3, 10]$$. Let's assume the correct interval is $$[3, 10]$$. Now, we will verify whether the function satisfies the conditions of LMVT on this interval. ### Checking the Continuity of $$f$$ on $$[3, 10]$$: The function is defined as a piecewise function: - For $$x < 1$$, $$f(x) = 0$$. However, since this does not affect the interval $$[3, 10]$$, it can be ignored. - For $$x \geq 1$$, $$f(x) = x$$. Since $$f(x) = x$$ is continuous on $$[3, 10]$$, $$f$$ is continuous on the interval. ### Checking the Differentiability of $$f$$ on $$(3, 10)$$: - For $$x \geq 1$$, the function $$f(x) = x$$ is differentiable everywhere on $$(3, 10)$$ because it’s a linear function. Thus, since $$f(x) = x$$ is both continuous on $$[3, 10]$$ and differentiable on $$(3, 10)$$, the LMVT is indeed applicable. ### Application of LMVT: According to LMVT, there exists a point $$c \in (3, 10)$$ such that: $$f'(c) = \frac{f(10) - f(3)}{10 - 3}.$$ Calculating the values: $$f(10) = 10 \quad \text{and} \quad f(3) = 3,$$ $$f'(x) = 1 \quad \text{for } x \geq 1,$$ $$f'(c) = \frac{10 - 3}{10 - 3} = 1.$$ Thus, there exists a $$c \in (3, 10)$$ such that $$f'(c) = 1$$, confirming the applicability of LMVT. --- Would you like further details or explanations? Here are some related questions to explore: 1. What happens if the function is not continuous at any point within the interval? 2. How would the Mean Value Theorem apply if the function was defined differently? 3. How does differentiability affect the application of LMVT? 4. What is the geometric interpretation of LMVT? 5. Can LMVT be applied to functions defined piecewise with discontinuities? **Tip:** For LMVT, ensure both continuity and differentiability conditions are met to apply the theorem correctly.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
Continuity
Differentiability
Piecewise Functions
Formulas
f'(c) = (f(b) - f(a)) / (b - a)
f(x) = 0 for x < 1; f(x) = x for x >= 1
Theorems
Lagrange Mean Value Theorem (LMVT)
Suitable Grade Level
Grades 11-12, College
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