Math Problem Statement
Find the value or values of c that satisfy the equation \(\frac{f(b) - f(a)}{b - a} = f'(c)\) in the conclusion of the Mean Value Theorem for the function and interval below.
\[g(x) = \begin{cases} x^3 - 2 & -2 \leq x \leq 0 \\ x^2 & 0 < x \leq 2 \end{cases}\]
Solution
To solve this problem, we need to apply the Mean Value Theorem (MVT) for the function over the interval .
Step 1: Verify Differentiability and Continuity
The function is defined piecewise:
\begin{cases} x^3 - 2 & \text{for } -2 \leq x \leq 0 \\ x^2 & \text{for } 0 < x \leq 2 \end{cases}$$ For the interval $$[0, 2]$$, we are only concerned with the part where $$g(x) = x^2$$, which is continuous and differentiable on $$[0, 2]$$. ### Step 2: Apply the Mean Value Theorem The MVT states that there exists a $$c$$ in the interval $$(0, 2)$$ such that $$g'(c) = \frac{g(2) - g(0)}{2 - 0}$$ #### Compute $$g(2)$$ and $$g(0)$$ - For $$g(2)$$: $$g(2) = 2^2 = 4$$ - For $$g(0)$$: $$g(0) = 0^2 = 0$$ Thus, $$\frac{g(2) - g(0)}{2 - 0} = \frac{4 - 0}{2} = 2$$ So, we need to find $$c$$ in $$(0, 2)$$ such that $$g'(c) = 2$$. ### Step 3: Differentiate $$g(x)$$ For $$0 < x \leq 2$$, $$g(x) = x^2$$. $$g'(x) = 2x$$ ### Step 4: Set Up the Equation We need $$g'(c) = 2$$, so $$2c = 2$$ Solving for $$c$$, $$c = 1$$ ### Conclusion The value of $$c$$ that satisfies the Mean Value Theorem for $$g(x)$$ on the interval $$[0, 2]$$ is $$c = 1$$. Would you like further details on any part of this solution? Here are some related questions for further exploration: 1. What is the Mean Value Theorem, and why is it applicable here? 2. How do you handle piecewise functions when applying the Mean Value Theorem? 3. Why is differentiability important in the Mean Value Theorem? 4. How would the solution change if the interval were $$[-2, 0]$$? 5. Can the Mean Value Theorem be applied to functions that are not differentiable on the entire interval? **Tip:** When applying the Mean Value Theorem, ensure the function is both continuous on the closed interval $$[a, b]$$ and differentiable on the open interval $$(a, b)$$.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
Mean Value Theorem
Piecewise Functions
Differentiation
Formulas
Mean Value Theorem: \(\frac{f(b) - f(a)}{b - a} = f'(c)\)
Derivative of \(x^2\): \(g'(x) = 2x\)
Theorems
Mean Value Theorem
Suitable Grade Level
Grades 11-12 or Calculus I
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