Math Problem Statement
Solution
The problem in the image asks to determine the values of and such that the given function is continuous for all real numbers. The function is defined piecewise as follows:
x^2 - 4 & \text{if } x < 2 \\ ax - b + 3 & \text{if } 2 \leq x < 3 \\ 2x - a & \text{if } x \geq 3 \end{cases}$$ ### Approach to solve: For the function to be continuous at $$x = 2$$ and $$x = 3$$, the left-hand limit and the right-hand limit at these points must equal the function values. #### Step 1: Continuity at $$x = 2$$ For $$f(x)$$ to be continuous at $$x = 2$$: 1. The left-hand limit as $$x \to 2^-$$ (from the first piece): $$\lim_{x \to 2^-} f(x) = 2^2 - 4 = 4 - 4 = 0$$ 2. The right-hand limit as $$x \to 2^+$$ (from the second piece): $$\lim_{x \to 2^+} f(x) = a(2) - b + 3 = 2a - b + 3$$ Since the function is continuous at $$x = 2$$, these limits must be equal: $$0 = 2a - b + 3$$ $$2a - b = -3 \quad \text{(Equation 1)}$$ #### Step 2: Continuity at $$x = 3$$ For $$f(x)$$ to be continuous at $$x = 3$$: 1. The left-hand limit as $$x \to 3^-$$ (from the second piece): $$\lim_{x \to 3^-} f(x) = a(3) - b + 3 = 3a - b + 3$$ 2. The right-hand limit as $$x \to 3^+$$ (from the third piece): $$\lim_{x \to 3^+} f(x) = 2(3) - a = 6 - a$$ Since the function is continuous at $$x = 3$$, these limits must be equal: $$3a - b + 3 = 6 - a$$ Simplifying this equation: $$3a - b + 3 = 6 - a$$ $$4a - b = 3 \quad \text{(Equation 2)}$$ #### Step 3: Solving the system of equations Now, we have the following system of equations: 1. $$2a - b = -3$$ 2. $$4a - b = 3$$ We can subtract Equation 1 from Equation 2 to eliminate $$b$$: $$(4a - b) - (2a - b) = 3 - (-3)$$ $$2a = 6$$ $$a = 3$$ Substitute $$a = 3$$ into Equation 1: $$2(3) - b = -3$$ $$6 - b = -3$$ $$b = 9$$ ### Final Answer: The values of $$a$$ and $$b$$ are: $$a = 3, \quad b = 9$$ Let me know if you have any questions or would like further clarification. --- Here are 5 related questions to expand your understanding: 1. What does it mean for a function to be continuous at a point? 2. How can we verify the continuity of a function at different points on a piecewise function? 3. What is the difference between a left-hand limit and a right-hand limit? 4. Can a function be continuous but not differentiable at some points? Why? 5. How does one solve a system of linear equations using substitution or elimination? **Tip:** Always check the left-hand and right-hand limits separately to ensure the continuity of a piecewise function at boundary points.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Limits
Piecewise Functions
System of Linear Equations
Formulas
Left-hand limit: lim_{x → c^-} f(x)
Right-hand limit: lim_{x → c^+} f(x)
System of linear equations solving methods
Theorems
For continuity at a point x = c, the left-hand limit and right-hand limit must be equal, and equal to f(c)
Suitable Grade Level
Grades 11-12
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