Math Problem Statement
Find the values of a and b that make the function f continuous everywhere. The function is defined as a piecewise function with three cases: (x^2 - 4)/(x - 2) if x < 2, ax^2 - bx + 3 if 2 <= x < 3, and 2x - a + b if x >= 3.
Solution
We are asked to find the values of and such that the piecewise function is continuous everywhere.
The function is defined as:
\begin{cases} \frac{x^2 - 4}{x - 2} & \text{if } x < 2 \\ ax^2 - bx + 3 & \text{if } 2 \leq x < 3 \\ 2x - a + b & \text{if } x \geq 3 \end{cases}$$ ### Step 1: Simplifying the first piece of the function For $$x < 2$$, the expression is $$\frac{x^2 - 4}{x - 2}$$. Notice that: $$x^2 - 4 = (x - 2)(x + 2)$$ So, the function simplifies to: $$\frac{(x - 2)(x + 2)}{x - 2} = x + 2 \quad \text{for} \quad x \neq 2$$ Thus, for $$x < 2$$, the function is $$f(x) = x + 2$$. ### Step 2: Ensuring continuity at $$x = 2$$ For the function to be continuous at $$x = 2$$, the limit of $$f(x)$$ as $$x$$ approaches 2 from both sides must be equal to $$f(2)$$. - As $$x \to 2^-$$ (from the left), $$f(x) = x + 2$$, so $$\lim_{x \to 2^-} f(x) = 2 + 2 = 4$$. - At $$x = 2$$, from the second piece $$f(x) = ax^2 - bx + 3$$. Substituting $$x = 2$$ into this gives: $$f(2) = a(2)^2 - b(2) + 3 = 4a - 2b + 3$$ To ensure continuity at $$x = 2$$, we set $$4a - 2b + 3 = 4$$. Simplifying: $$4a - 2b = 1 \quad \text{(Equation 1)}$$ ### Step 3: Ensuring continuity at $$x = 3$$ For the function to be continuous at $$x = 3$$, the limit of $$f(x)$$ as $$x$$ approaches 3 from both sides must also be equal. - As $$x \to 3^-$$ (from the left), $$f(x) = ax^2 - bx + 3$$. Substituting $$x = 3$$ gives: $$f(3) = a(3)^2 - b(3) + 3 = 9a - 3b + 3$$ - As $$x \to 3^+$$ (from the right), $$f(x) = 2x - a + b$$. Substituting $$x = 3$$ gives: $$f(3) = 2(3) - a + b = 6 - a + b$$ To ensure continuity at $$x = 3$$, set $$9a - 3b + 3 = 6 - a + b$$. Simplifying this: $$9a - 3b + 3 = 6 - a + b$$ $$10a - 4b = 3 \quad \text{(Equation 2)}$$ ### Step 4: Solving the system of equations We now solve the system of two equations: 1. $$4a - 2b = 1$$ 2. $$10a - 4b = 3$$ We can solve this system using substitution or elimination. I'll use elimination here. Multiply the first equation by 2: $$8a - 4b = 2$$ Now subtract the second equation from this: $$(8a - 4b) - (10a - 4b) = 2 - 3$$ $$-2a = -1 \quad \Rightarrow \quad a = \frac{1}{2}$$ Substitute $$a = \frac{1}{2}$$ into the first equation: $$4\left(\frac{1}{2}\right) - 2b = 1$$ $$2 - 2b = 1 \quad \Rightarrow \quad -2b = -1 \quad \Rightarrow \quad b = \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}$$ --- Would you like more details or have any questions about the steps? Here are five related questions to explore: 1. How does one check differentiability for a piecewise function? 2. Can you explain how limits work when verifying continuity? 3. What are common techniques for solving systems of linear equations? 4. What are the conditions required for a function to be continuous? 5. How would the process change if $$f(x)$$ were defined differently for one of the intervals? **Tip**: Always check for removable discontinuities when dealing with rational functions!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
Formulas
Continuity conditions at a point: lim(x->c-) f(x) = lim(x->c+) f(x) = f(c)
Factoring quadratics
Theorems
Definition of continuity
Limit laws
Solving systems of linear equations
Suitable Grade Level
Grades 11-12
Related Recommendation
Find a and b for Continuity of a Piecewise Function Involving Quadratics
Find Values of a and b for Continuity in Piecewise Function
Find values of a and b for continuity in piecewise function
Find values of a and b for continuity of piecewise function at x = 1
Finding a and b for Continuity of Piecewise Function with Limits