Math Problem Statement
what values of a and b make f(x) continous everywhere for f(x)={x^2-4/x-2 for x<2, ax^2-bx-30 for 2<=x<3 16x-a+b for 3<=x
Solution
To find the values of and that make the function continuous everywhere, we need to ensure continuity at the points and . Specifically, the function must have the same value from both the left and right at these points.
The function is defined piecewise as follows:
\frac{x^2 - 4}{x - 2} & \text{for } x < 2, \\ ax^2 - bx - 30 & \text{for } 2 \leq x < 3, \\ 16x - a + b & \text{for } 3 \leq x. \end{cases}$$ ### Step 1: Continuity at $$x = 2$$ For the function to be continuous at $$x = 2$$, the limit from the left ($$x \to 2^{-}$$) must equal the limit from the right ($$x \to 2^{+}$$), and the function must be defined at $$x = 2$$. #### Left-hand limit as $$x \to 2^-$$: For $$x < 2$$, we have $$f(x) = \frac{x^2 - 4}{x - 2}$$. Simplifying this expression: $$\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2 \quad \text{for } x \neq 2.$$ So, as $$x \to 2^-$$, the limit of $$f(x)$$ is: $$\lim_{x \to 2^-} f(x) = 2 + 2 = 4.$$ #### Right-hand limit as $$x \to 2^+$$: For $$x \geq 2$$, we have $$f(x) = ax^2 - bx - 30$$. To ensure continuity at $$x = 2$$, we require: $$a(2)^2 - b(2) - 30 = 4.$$ Simplifying: $$4a - 2b - 30 = 4.$$ Thus, the first equation is: $$4a - 2b = 34. \quad \text{(Equation 1)}$$ ### Step 2: Continuity at $$x = 3$$ For the function to be continuous at $$x = 3$$, the limit from the left ($$x \to 3^{-}$$) must equal the limit from the right ($$x \to 3^{+}$$). #### Left-hand limit as $$x \to 3^-$$: For $$2 \leq x < 3$$, we have $$f(x) = ax^2 - bx - 30$$. So, as $$x \to 3^{-}$$, the limit of $$f(x)$$ is: $$a(3)^2 - b(3) - 30 = 9a - 3b - 30.$$ #### Right-hand limit as $$x \to 3^+$$: For $$x \geq 3$$, we have $$f(x) = 16x - a + b$$. So, as $$x \to 3^{+}$$, the limit of $$f(x)$$ is: $$16(3) - a + b = 48 - a + b.$$ To ensure continuity at $$x = 3$$, we require: $$9a - 3b - 30 = 48 - a + b.$$ Simplifying: $$9a - 3b + a - b = 78.$$ Thus, the second equation is: $$10a - 4b = 78. \quad \text{(Equation 2)}$$ ### Step 3: Solving the system of equations We now have the following system of equations: 1. $$4a - 2b = 34$$, 2. $$10a - 4b = 78$$. To solve this system, we can use substitution or elimination. I'll solve it using the elimination method. #### Eliminate $$b$$: Multiply the first equation by 2: $$8a - 4b = 68.$$ Now subtract this equation from the second equation: $$(10a - 4b) - (8a - 4b) = 78 - 68,$$ which simplifies to: $$2a = 10.$$ Thus, $$a = 5$$. #### Solve for $$b$$: Substitute $$a = 5$$ into the first equation: $$4(5) - 2b = 34,$$ which simplifies to: $$20 - 2b = 34,$$ so $$2b = -14$$, and therefore $$b = -7$$. ### Final Answer: The values of $$a$$ and $$b$$ that make $$f(x)$$ continuous everywhere are: $$a = 5, \quad b = -7.$$ Would you like more details on the solution, or do you have any questions? Here are 5 related questions: 1. How do you check for continuity at other points in a piecewise function? 2. What is the general procedure for solving systems of equations? 3. How does differentiability relate to continuity in piecewise functions? 4. What happens if a function is not continuous at a point? 5. Can a function be continuous but not differentiable? **Tip**: When solving piecewise functions, always check for both limits and equality of function values at the boundary points.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Continuity
Limits
Solving Systems of Equations
Formulas
Limit: \lim_{x \to c} f(x)
System of linear equations: 4a - 2b = 34, 10a - 4b = 78
Theorems
Continuity at a Point
Limit Theorem
Suitable Grade Level
Grades 10-12
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