Math Problem Statement
resuelve
Determine the constants A and B so that the function f given below is continuous throughout its domain.
f(x) = { (x^2 - 4) / (x - 2), if x < 2; Ax^2 - Bx + 3, if 2 ≤ x < 3; 2x - A + B, if x ≥ 3 }
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Solution
To ensure that the function is continuous across its entire domain, we need to find the constants and such that there are no discontinuities at the points where the function definition changes, namely at and .
The function is given by:
\frac{x^2 - 4}{x - 2}, & x < 2 \\ Ax^2 - Bx + 3, & 2 \leq x < 3 \\ 2x - A + B, & x \geq 3 \end{cases}$$ ### Step 1: Simplify the expression for $$f(x)$$ when $$x < 2$$ For $$x < 2$$, we have: $$f(x) = \frac{x^2 - 4}{x - 2}$$ This expression can be simplified by factoring $$x^2 - 4$$ as $$(x + 2)(x - 2)$$: $$f(x) = \frac{(x + 2)(x - 2)}{x - 2} = x + 2, \quad x \neq 2$$ So, for $$x < 2$$, $$f(x) = x + 2$$. ### Step 2: Continuity at $$x = 2$$ To ensure continuity at $$x = 2$$, the limit from the left as $$x \to 2^-$$ must equal the value from the right as $$x \to 2^+$$. 1. **Limit as $$x \to 2^-$$:** $$\lim_{x \to 2^-} f(x) = 2 + 2 = 4$$ 2. **Value at $$x = 2$$ (from the middle function):** $$f(2) = A(2)^2 - B(2) + 3 = 4A - 2B + 3$$ To ensure continuity at $$x = 2$$: $$4A - 2B + 3 = 4$$ Simplifying, we get: $$4A - 2B = 1 \quad \Rightarrow \quad 2A - B = \frac{1}{2}$$ ### Step 3: Continuity at $$x = 3$$ To ensure continuity at $$x = 3$$, the limit from the left as $$x \to 3^-$$ must equal the value from the right as $$x \to 3^+$$. 1. **Limit as $$x \to 3^-$$:** $$\lim_{x \to 3^-} f(x) = A(3)^2 - B(3) + 3 = 9A - 3B + 3$$ 2. **Value at $$x = 3$$ (from the last function):** $$f(3) = 2(3) - A + B = 6 - A + B$$ To ensure continuity at $$x = 3$$: $$9A - 3B + 3 = 6 - A + B$$ Simplifying, we get: $$10A - 4B = 3 \quad \Rightarrow \quad 5A - 2B = \frac{3}{2}$$ ### Step 4: Solve the system of equations We now have the following system of equations: 1. $$2A - B = \frac{1}{2}$$ 2. $$5A - 2B = \frac{3}{2}$$ Solving this system will give us the values of $$A$$ and $$B$$ that make $$f(x)$$ continuous across its entire domain. Let me calculate these values.The solution to the system of equations is: $$A = 0.5 \quad \text{and} \quad B = 0.5$$ Thus, for the function $$f(x)$$ to be continuous across its entire domain, the constants $$A$$ and $$B$$ must both be equal to 0.5. Would you like further details on this solution, or do you have any questions? Here are five related questions to deepen understanding: 1. How would you verify that $$f(x)$$ is continuous at all points with the found values of $$A$$ and $$B$$? 2. What happens if different values for $$A$$ and $$B$$ are chosen? Would the function still be continuous? 3. Can you generalize this process for ensuring continuity in piecewise functions? 4. How would differentiability affect the choice of $$A$$ and $$B$$ for this function? 5. Could you apply the same steps if there were additional intervals in the piecewise function? **Tip:** When dealing with piecewise functions, always simplify each piece as much as possible before setting up equations for continuity or differentiability.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Piecewise Functions
Limits
Formulas
Limit definition: lim(x→c) f(x) = f(c)
Factoring quadratic expressions
Solving systems of equations
Theorems
Continuity at a Point
Suitable Grade Level
Grades 10-12