Math Problem Statement
Find the values of a and b that make f(x) continuous and differentiable at x = 1. The function is given as a piecewise function with f(x) = x^2 - 2x - a for x < 1, and f(x) = -x^2 + bx + 2 for x >= 1.
Solution
We are given a piecewise function :
\begin{cases} x^2 - 2x - a & \text{if } x < 1, \\ -x^2 + bx + 2 & \text{if } x \geq 1, \end{cases}$$ and we need to find the values of $$a$$ and $$b$$ such that $$f(x)$$ is continuous and differentiable at $$x = 1$$. ### Step 1: Ensure Continuity at $$x = 1$$ For $$f(x)$$ to be continuous at $$x = 1$$, the left-hand limit and right-hand limit must be equal to $$f(1)$$. This means: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1).$$ 1. From the first piece of the function, when $$x < 1$$: $$\lim_{x \to 1^-} f(x) = 1^2 - 2(1) - a = 1 - 2 - a = -1 - a.$$ 2. From the second piece of the function, when $$x \geq 1$$: $$f(1) = -(1)^2 + b(1) + 2 = -1 + b + 2 = b + 1.$$ To ensure continuity, we set the limits equal to each other: $$-1 - a = b + 1.$$ This simplifies to: $$a + b = -2. \quad \text{(Equation 1)}$$ ### Step 2: Ensure Differentiability at $$x = 1$$ For $$f(x)$$ to be differentiable at $$x = 1$$, the derivatives from both sides must be equal at $$x = 1$$. So we first find the derivatives of the two pieces of the function. 1. For $$x < 1$$, $$f(x) = x^2 - 2x - a$$. The derivative is: $$f'(x) = 2x - 2.$$ Evaluating at $$x = 1$$: $$f'(1^-) = 2(1) - 2 = 0.$$ 2. For $$x \geq 1$$, $$f(x) = -x^2 + bx + 2$$. The derivative is: $$f'(x) = -2x + b.$$ Evaluating at $$x = 1$$: $$f'(1^+) = -2(1) + b = -2 + b.$$ To ensure differentiability, we set the derivatives equal at $$x = 1$$: $$0 = -2 + b.$$ This simplifies to: $$b = 2. \quad \text{(Equation 2)}$$ ### Step 3: Solve for $$a$$ Now substitute $$b = 2$$ into Equation 1: $$a + 2 = -2,$$ which gives: $$a = -4.$$ ### Final Answer The values of $$a$$ and $$b$$ that make $$f(x)$$ continuous and differentiable at $$x = 1$$ are: $$a = -4, \quad b = 2.$$ Would you like a more detailed breakdown of any steps? --- Here are 5 related questions for further exploration: 1. What are the necessary conditions for a function to be continuous at a point? 2. How can you check the differentiability of a function at a point with a more complex piecewise function? 3. What happens if a function is continuous but not differentiable at a point? 4. How can the chain rule be used when differentiating piecewise functions with more complex boundaries? 5. How do second-order derivatives behave in piecewise functions at boundary points? **Tip:** When dealing with piecewise functions, always ensure to check both continuity and differentiability by comparing limits and derivatives from both sides of the boundary.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Continuity
Differentiability
Piecewise Functions
Quadratic Functions
Formulas
Continuity Condition: lim(x→1^-) f(x) = lim(x→1^+) f(x)
Differentiability Condition: f'(1^-) = f'(1^+)
Theorems
Continuity Theorem
Differentiability Theorem
Suitable Grade Level
Grade 11-12
Related Recommendation
Finding Constants for Continuity and Differentiability in a Piecewise Function
Continuity and Differentiability of a Piecewise Function at a Point
Finding a and b for Continuity in a Piecewise Function
Find a and b for Continuity and Differentiability in a Piecewise Function
Solving for Continuity in a Piecewise Function