Math Problem Statement
Tentukanlah konstanta a dan b agar fungsi f(x) kontinu di selang [0, ∞): f(x) = {ax^2 + b / x^2 - 4, 0 ≤ x < 2; x^2 - 3x + 5, x ≥ 2}.
Solution
The problem asks to determine the constants and so that the function is continuous over the interval .
The function is defined piecewise as:
\frac{ax^2 + b}{x^2 - 4}, & 0 \leq x < 2 \\ x^2 - 3x + 5, & x \geq 2 \end{cases}$$ To ensure the function is continuous at $$x = 2$$, the two pieces of the function must have the same value at $$x = 2$$. This gives us a condition for continuity. ### Step 1: Find $$f(2)$$ for both expressions 1. **For $$x \geq 2$$:** $$f(2) = 2^2 - 3(2) + 5 = 4 - 6 + 5 = 3$$ 2. **For $$0 \leq x < 2$$:** $$f(2) = \frac{a(2)^2 + b}{(2)^2 - 4} = \frac{4a + b}{4 - 4} = \frac{4a + b}{0}$$ Since dividing by zero is undefined, the numerator must also be zero for the expression to be valid. This gives us the condition: $$4a + b = 0$$ (Equation 1) ### Step 2: Apply continuity condition We know from the other piece of the function that at $$x = 2$$, $$f(2) = 3$$. Thus: $$f(2) = 3 \quad \text{for both pieces}.$$ Therefore, the limit as $$x$$ approaches 2 from the left must also be 3. Using the first part of the piecewise function: $$f(2) = 3 = \frac{4a + b}{0} \quad \text{(from left, at $$x = 2$$)}$$ We already know that $$4a + b = 0$$, so from Equation 1: $$4a + b = 0$$ ### Step 3: Solve for $$a$$ and $$b$$Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity of functions
Piecewise functions
Solving for constants
Formulas
Continuity condition: f(a) = lim_{x → a⁻} f(x) = lim_{x → a⁺} f(x)
Piecewise function evaluation
Solving linear equations (4a + b = 0)
Theorems
Continuity theorem at a point
Suitable Grade Level
University level (First-year calculus or pre-calculus)