Math Problem Statement
Tentukan nilai a dan b agar f(x) = \left(\frac{ax^3 + b}{x^2 - 4}, \; 0 \leq x < 2 \; \text{and} \; x^3 - 3x + 1, \; x \geq 2 \right) kontinu di [0, \infty).
Solution
The problem here requires us to find the values of and to make the function continuous on the interval .
The function is defined as:
\frac{a x^3 + b}{x^2 - 4}, & 0 \leq x < 2, \\ x^3 - 3x + 1, & x \geq 2 \end{cases}$$ For $$f(x)$$ to be continuous at $$x = 2$$, the left-hand limit as $$x$$ approaches 2 must be equal to the right-hand limit as $$x$$ approaches 2, and both must be equal to $$f(2)$$. ### Step-by-Step Solution: 1. **Evaluate $$f(2)$$:** $$f(2) = 2^3 - 3 \cdot 2 + 1 = 8 - 6 + 1 = 3.$$ Therefore, $$f(x)$$ must approach 3 from both sides as $$x \to 2$$. 2. **Left-Hand Limit as $$x \to 2^-$$:** We have: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{a x^3 + b}{x^2 - 4}.$$ Notice that $$x^2 - 4 = (x - 2)(x + 2)$$. When $$x \to 2$$, the denominator approaches zero, so we need the numerator to also approach zero for the limit to be finite. Set up the condition: $$a \cdot 2^3 + b = 0.$$ This gives: $$8a + b = 0 \quad \Rightarrow \quad b = -8a.$$ 3. **Substitute $$b = -8a$$ and Calculate the Limit:** Substitute $$b = -8a$$ into the left-hand limit: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{a x^3 - 8a}{(x - 2)(x + 2)}.$$ Factor the numerator: $$= \lim_{x \to 2^-} \frac{a (x^3 - 8)}{(x - 2)(x + 2)} = \lim_{x \to 2^-} \frac{a (x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)}.$$ Cancel $$x - 2$$ from the numerator and denominator: $$= \lim_{x \to 2^-} \frac{a (x^2 + 2x + 4)}{x + 2}.$$ Substitute $$x = 2$$: $$= \frac{a (2^2 + 2 \cdot 2 + 4)}{2 + 2} = \frac{a (4 + 4 + 4)}{4} = \frac{12a}{4} = 3a.$$ For continuity, this limit must equal $$f(2) = 3$$: $$3a = 3 \quad \Rightarrow \quad a = 1.$$ 4. **Find $$b$$:** Substitute $$a = 1$$ into $$b = -8a$$: $$b = -8 \cdot 1 = -8.$$ ### Final Answer: The values of $$a$$ and $$b$$ that make $$f(x)$$ continuous on $$[0, \infty)$$ are: $$a = 1, \quad b = -8.$$ Would you like further details or have any questions about this solution? --- Here are some related questions to explore further: 1. What is the definition of continuity for piecewise functions? 2. How do we handle removable discontinuities in rational functions? 3. What is the significance of factoring when evaluating limits involving zero denominators? 4. How would the solution change if the piecewise condition had a different breakpoint? 5. How does the factorization of polynomials help simplify limits? **Tip:** Always check both the left-hand and right-hand limits for continuity, especially at points where the function's definition changes.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Continuity
Piecewise Functions
Limits
Polynomials
Factoring
Formulas
Continuity condition: \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)
Factoring: x^2 - 4 = (x - 2)(x + 2)
Theorems
Continuity of piecewise functions
Suitable Grade Level
Grade 11-12
Related Recommendation
Determine Continuity of Piecewise Function at x = 1
Finding Continuity of a Piecewise Function Involving Trigonometric and Polynomial Functions
Continuity of Piecewise Function at x = 0
Determine Constants for Continuity in a Piecewise Function
Find values of a and b for continuity in a piecewise function