Math Problem Statement
Solution
The problem asks to find the values of and that make the given function continuous everywhere. The function is defined in piecewise form as:
\begin{cases} \frac{x^2 - 4}{x - 2} & \text{if } x < 2 \\ Ax^2 - Bx + 3 & \text{if } 2 \leq x \leq 3 \\ 2x - A + B & \text{if } x \geq 3 \end{cases}$$ ### Step 1: Continuity at $$x = 2$$ To ensure continuity at $$x = 2$$, the left-hand limit and the right-hand limit must be equal at that point. - For $$x < 2$$, we simplify the expression $$\frac{x^2 - 4}{x - 2}$$: $$\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2, \text{ for } x \neq 2.$$ So as $$x \to 2^-$$, $$f(x) \to 4$$. - For $$x = 2$$, using the second part $$f(x) = Ax^2 - Bx + 3$$: $$f(2) = A(2)^2 - B(2) + 3 = 4A - 2B + 3.$$ To ensure continuity, we set the left-hand and right-hand limits equal at $$x = 2$$: $$4A - 2B + 3 = 4.$$ Simplifying this, we get: $$4A - 2B = 1. \tag{1}$$ ### Step 2: Continuity at $$x = 3$$ For $$x = 3$$, we again match the left-hand and right-hand limits: - For $$x = 3$$, using the second part $$Ax^2 - Bx + 3$$: $$f(3) = A(3)^2 - B(3) + 3 = 9A - 3B + 3.$$ - For $$x = 3$$, using the third part $$f(x) = 2x - A + B$$: $$f(3) = 2(3) - A + B = 6 - A + B.$$ To ensure continuity, set the left-hand and right-hand limits equal at $$x = 3$$: $$9A - 3B + 3 = 6 - A + B.$$ Simplifying this equation: $$10A - 4B = 3. \tag{2}$$ ### Step 3: Solve the System of Equations Now we have the system of equations: 1. $$4A - 2B = 1$$ 2. $$10A - 4B = 3$$ Solve equation (1) for $$B$$: $$2B = 4A - 1 \implies B = 2A - \frac{1}{2}.$$ Substitute this into equation (2): $$10A - 4(2A - \frac{1}{2}) = 3.$$ Simplify: $$10A - 8A + 2 = 3 \implies 2A + 2 = 3 \implies 2A = 1 \implies A = \frac{1}{2}.$$ Now substitute $$A = \frac{1}{2}$$ back into $$B = 2A - \frac{1}{2}$$: $$B = 2\left(\frac{1}{2}\right) - \frac{1}{2} = 1 - \frac{1}{2} = \frac{1}{2}.$$ ### Final Answer: The values of $$A$$ and $$B$$ that make the function continuous everywhere are: $$A = \frac{1}{2}, \quad B = \frac{1}{2}.$$ Would you like further details or clarifications? Here are some related questions for practice: 1. How do you determine if a piecewise function is continuous at a given point? 2. What happens if the function is not continuous at a point? 3. Can a function be continuous but not differentiable? 4. How would you adjust this problem if the intervals were different? 5. What is the significance of limits when analyzing piecewise functions? **Tip:** Always check both the left-hand and right-hand limits to ensure continuity of piecewise functions at the given points.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Piecewise Functions
Limits
Algebra
Formulas
Left-hand and right-hand limits for continuity: lim_{x -> c^-} f(x) = lim_{x -> c^+} f(x)
Factoring quadratics: (x^2 - 4)/(x - 2) = x + 2 for x ≠ 2
Theorems
Continuity at a point: A function is continuous at x = c if the left-hand limit, right-hand limit, and the value of the function at c are equal.
Solving systems of linear equations
Suitable Grade Level
Grade 11-12
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