Math Problem Statement
For what values of a, m, and b does the function f(x) satisfy the hypotheses of the mean value theorem on the interval left bracket 0 comma 3 right bracket? f(x)equalsleft brace Start 3 By 3 Matrix 1st Row 1st Column negative 4 2nd Column 3rd Column x equals 0 2nd Row 1st Column negative x squared plus 2 x plus a 2nd Column 3rd Column 0 less than x less than 2 3rd Row 1st Column mx plus b 2nd Column 3rd Column 2 less than or equals x less than or equals 3 EndMatrix
Solution
To determine the values of , , and that make satisfy the hypotheses of the Mean Value Theorem (MVT) on the interval , we need to verify that:
- Continuity on the closed interval .
- Differentiability on the open interval .
The function is defined in three parts:
-4, & x = 0 \\ -x^2 + 2x + a, & 0 < x < 2 \\ mx + b, & 2 \leq x \leq 3 \end{cases}$$ To satisfy the continuity requirement, $$f(x)$$ must be continuous at $$x = 0$$, $$x = 2$$, and $$x = 3$$. We’ll address each of these points individually. ### Step 1: Continuity at $$x = 0$$ For $$f(x)$$ to be continuous at $$x = 0$$: $$\lim_{x \to 0^+} f(x) = f(0)$$ Since $$f(0) = -4$$, we need: $$\lim_{x \to 0^+} (-x^2 + 2x + a) = -4$$ Calculating the limit: $$\lim_{x \to 0^+} (-x^2 + 2x + a) = a$$ Thus, $$a = -4$$ to ensure continuity at $$x = 0$$. ### Step 2: Continuity at $$x = 2$$ For $$f(x)$$ to be continuous at $$x = 2$$: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)$$ The left-hand limit as $$x \to 2^-$$ is: $$\lim_{x \to 2^-} (-x^2 + 2x + a) = -2^2 + 2 \cdot 2 + a = -4 + 4 + a = a$$ Since $$a = -4$$, the left-hand limit at $$x = 2$$ is $$-4$$. The right-hand limit as $$x \to 2^+$$ is: $$\lim_{x \to 2^+} (mx + b) = 2m + b$$ To ensure continuity at $$x = 2$$, we set: $$a = 2m + b$$ Since $$a = -4$$, this implies: $$-4 = 2m + b$$ ### Step 3: Continuity at $$x = 3$$ For $$f(x)$$ to be continuous at $$x = 3$$: $$f(3) = m \cdot 3 + b = 3m + b$$ There is no additional condition on $$f(3)$$ for continuity here, but we’ll keep $$3m + b$$ as the expression for $$f(3)$$. ### Step 4: Differentiability at $$x = 2$$ For $$f(x)$$ to be differentiable at $$x = 2$$, the derivatives from both sides must match at $$x = 2$$. The derivative of $$f(x) = -x^2 + 2x + a$$ for $$0 < x < 2$$ is: $$f'(x) = -2x + 2$$ Thus, $$f'(2) = -2 \cdot 2 + 2 = -4 + 2 = -2$$. The derivative of $$f(x) = mx + b$$ for $$2 < x \leq 3$$ is: $$f'(x) = m$$ For differentiability at $$x = 2$$, we need: $$m = -2$$ ### Step 5: Solve for $$b$$ With $$m = -2$$, we substitute back into the continuity condition at $$x = 2$$: $$-4 = 2(-2) + b \Rightarrow -4 = -4 + b$$ Thus, $$b = 0$$. ### Conclusion The values of $$a$$, $$m$$, and $$b$$ that satisfy the hypotheses of the Mean Value Theorem on the interval $$[0, 3]$$ are: $$a = -4, \quad m = -2, \quad b = 0$$ Would you like further explanation on any part of the process? 1. How does differentiability at a point ensure the application of the Mean Value Theorem? 2. Can the Mean Value Theorem apply if only continuity is met but not differentiability? 3. Why is checking both sides of each boundary point crucial for continuity in piecewise functions? 4. What are common functions that automatically satisfy the MVT on any interval? 5. How would different values of $$a$$, $$m$$, or $$b$$ impact the continuity or differentiability in this example? **Tip:** The Mean Value Theorem applies only if both continuity and differentiability conditions are strictly met within the given interval.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Mean Value Theorem
Continuity
Differentiability
Piecewise Functions
Formulas
Continuity condition: lim(x -> c-) f(x) = lim(x -> c+) f(x)
Differentiability condition: f'(x) from left = f'(x) from right
Theorems
Mean Value Theorem
Suitable Grade Level
Grades 11-12
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