Math Problem Statement
solve it
Solution
To determine whether Lagrange's Mean Value Theorem (LMVT) is applicable for the given piecewise function on the interval , we must check two conditions:
- Continuity: The function must be continuous on the closed interval .
- Differentiability: The function must be differentiable on the open interval .
The function is defined as:
\begin{cases} 2 + x^2 & \text{if } x \leq 1, \\ 3x & \text{if } x > 1. \end{cases}$$ ### Step 1: Verify Continuity on $$[-1, 2]$$ - **For $$x \leq 1$$:** $$f(x) = 2 + x^2$$ is a polynomial, so it is continuous for all $$x$$, including $$x = 1$$. - **For $$x > 1$$:** $$f(x) = 3x$$ is also a polynomial, so it is continuous for all $$x$$, including $$x = 2$$. - **At $$x = 1$$:** Check whether the left-hand limit (LHL) equals the right-hand limit (RHL) and the value of $$f(1)$$: - LHL as $$x \to 1^-$$: $$f(x) = 2 + x^2$$, so $$\text{LHL} = 2 + 1^2 = 3$$. - RHL as $$x \to 1^+$$: $$f(x) = 3x$$, so $$\text{RHL} = 3(1) = 3$$. - $$f(1)$$: By definition of $$f(x)$$, when $$x \leq 1$$, $$f(1) = 2 + 1^2 = 3$$. Since LHL = RHL = $$f(1)$$, $$f(x)$$ is continuous at $$x = 1$$. Thus, $$f(x)$$ is continuous on $$[-1, 2]$$. ### Step 2: Verify Differentiability on $$(-1, 2)$$ - **For $$x < 1$$:** $$f(x) = 2 + x^2$$, so $$f'(x) = 2x$$. Since $$f'(x)$$ exists for all $$x$$ in this range, $$f(x)$$ is differentiable for $$x < 1$$. - **For $$x > 1$$:** $$f(x) = 3x$$, so $$f'(x) = 3$$. Since $$f'(x)$$ exists for all $$x$$ in this range, $$f(x)$$ is differentiable for $$x > 1$$. - **At $$x = 1$$:** Check whether the derivative from the left equals the derivative from the right: - Left-hand derivative (LHD): For $$x \to 1^-$$, $$f'(x) = 2x$$, so $$\text{LHD} = 2(1) = 2$$. - Right-hand derivative (RHD): For $$x \to 1^+$$, $$f'(x) = 3$$, so $$\text{RHD} = 3$$. Since LHD $$\neq$$ RHD at $$x = 1$$, $$f(x)$$ is **not differentiable** at $$x = 1$$. ### Step 3: Conclusion on Applicability of LMVT Lagrange's Mean Value Theorem requires that $$f(x)$$ is differentiable on the open interval $$(-1, 2)$$. However, $$f(x)$$ is not differentiable at $$x = 1$$. Therefore, **LMVT is not applicable** for the function $$f(x)$$ on the interval $$[-1, 2]$$. --- Let me know if you'd like a deeper explanation or further examples! Here are 5 related questions: 1. What are the precise conditions required for the Mean Value Theorem to hold? 2. Can a function be continuous but not differentiable? Provide examples. 3. How does LMVT differ from Rolle's Theorem? 4. Can a function be differentiable but not continuous? Why or why not? 5. How would the theorem apply if $$f(x)$$ were differentiable on the entire interval? **Tip:** To apply LMVT successfully, always check continuity and differentiability thoroughly across the given interval.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Continuity
Differentiability
Lagrange's Mean Value Theorem (LMVT)
Formulas
Derivative of f(x)
Continuity condition: Limit from the left = Limit from the right = f(value)
Theorems
Lagrange's Mean Value Theorem (LMVT)
Suitable Grade Level
Grades 11-12
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