Math Problem Statement
f(x) =
x2 − 4 x − 2 if x < 2 ax2 − bx + 3 if 2 ≤ x < 3 2x − a + b if x ≥ 3
Solution
We are given a piecewise function , defined as follows:
\begin{cases} \frac{x^2 - 4}{x - 2}, & \text{if } x < 2, \\ ax^2 - bx + 3, & \text{if } 2 \leq x < 3, \\ 2x - a + b, & \text{if } x \geq 3. \end{cases}$$ This function is composed of three distinct parts for different intervals of $$x$$. To analyze the function, we often need to ensure continuity and/or differentiability at the points where the intervals meet ($$x = 2$$ and $$x = 3$$). ### Step 1: Simplify $$f(x)$$ for $$x < 2$$ The first case for $$x < 2$$ is: $$\frac{x^2 - 4}{x - 2}.$$ Factorizing the numerator, we get: $$x^2 - 4 = (x - 2)(x + 2).$$ Thus, $$\frac{x^2 - 4}{x - 2} = x + 2 \quad \text{for } x \neq 2.$$ So the function simplifies to $$f(x) = x + 2$$ for $$x < 2$$. --- ### Step 2: Check for continuity at $$x = 2$$ For $$f(x)$$ to be continuous at $$x = 2$$, the left-hand limit ($$\lim_{x \to 2^-}$$), the right-hand limit ($$\lim_{x \to 2^+}$$), and the value of $$f(2)$$ must all be equal. #### Left-hand limit as $$x \to 2^-$$: Using $$f(x) = x + 2$$ for $$x < 2$$, $$\lim_{x \to 2^-} f(x) = 2 + 2 = 4.$$ #### Right-hand limit as $$x \to 2^+$$: For $$2 \leq x < 3$$, $$f(x) = ax^2 - bx + 3$$. Substituting $$x = 2$$, $$\lim_{x \to 2^+} f(x) = a(2^2) - b(2) + 3 = 4a - 2b + 3.$$ To ensure continuity at $$x = 2$$, these limits must be equal: $$4 = 4a - 2b + 3.$$ Simplifying: $$4a - 2b = 1. \tag{1}$$ --- ### Step 3: Check for continuity at $$x = 3$$ For $$f(x)$$ to be continuous at $$x = 3$$, the left-hand limit ($$\lim_{x \to 3^-}$$) and the right-hand limit ($$\lim_{x \to 3^+}$$) must be equal. #### Left-hand limit as $$x \to 3^-$$: For $$2 \leq x < 3$$, $$f(x) = ax^2 - bx + 3$$. Substituting $$x = 3$$, $$\lim_{x \to 3^-} f(x) = a(3^2) - b(3) + 3 = 9a - 3b + 3.$$ #### Right-hand limit as $$x \to 3^+$$: For $$x \geq 3$$, $$f(x) = 2x - a + b$$. Substituting $$x = 3$$, $$\lim_{x \to 3^+} f(x) = 2(3) - a + b = 6 - a + b.$$ To ensure continuity at $$x = 3$$, these limits must be equal: $$9a - 3b + 3 = 6 - a + b.$$ Simplifying: $$10a - 4b = 3. \tag{2}$$ --- ### Step 4: Solve for $$a$$ and $$b$$ We now solve the system of equations: 1. $$4a - 2b = 1$$, 2. $$10a - 4b = 3$$. Divide the second equation by 2: $$5a - 2b = 1.5. \tag{3}$$ Subtract equation (1) from (3): $$(5a - 2b) - (4a - 2b) = 1.5 - 1,$$ $$a = 0.5.$$ Substitute $$a = 0.5$$ into equation (1): $$4(0.5) - 2b = 1,$$ $$2 - 2b = 1,$$ $$b = 0.5.$$ --- ### Final Values: $$a = 0.5, \quad b = 0.5.$$ --- Would you like me to verify the solution or explain further? 😊 ### 5 Related Questions: 1. What does it mean for a function to be continuous at a point? 2. How do we check differentiability for a piecewise function? 3. Can you simplify rational expressions like $$\frac{x^2 - 4}{x - 2}$$ for any $$x$$? 4. What happens if $$a$$ and $$b$$ do not satisfy continuity conditions? 5. How does continuity differ from differentiability in a piecewise function? ### 1 Tip: When solving piecewise functions, always check both continuity and differentiability at the boundary points for complete analysis!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Continuity
Limits
Algebra
Formulas
f(x) = x + 2 for x < 2
f(x) = ax^2 - bx + 3 for 2 ≤ x < 3
f(x) = 2x - a + b for x ≥ 3
Theorems
Continuity at a point
Limit laws
Suitable Grade Level
Grades 11-12