Math Problem Statement
Find the values of a and b such that the piecewise function f(x) is continuous at x = 1.
Solution
The problem you're dealing with seems to ask for continuity of the function at , where the function is piecewise-defined. Let's break it down:
\begin{cases} 3x^2 + 2ax - 1, & x > 1 \\ 4a, & x = 1 \\ \frac{3b \sin(x-1)}{x-1} - 2, & x < 1 \end{cases}$$ The condition given is that $$f(x)$$ must be continuous at $$x = 1$$. For continuity at $$x = 1$$, the following conditions must be met: 1. $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$$. Let's compute each limit and set up the equations to solve for $$a$$ and $$b$$. ### 1. $$\lim_{x \to 1^-} f(x)$$: When $$x < 1$$, the function is given as $$f(x) = \frac{3b \sin(x-1)}{x-1} - 2$$. As $$x \to 1$$, we can apply the known limit $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$. Therefore: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \left( \frac{3b \sin(x-1)}{x-1} - 2 \right) = 3b - 2.$$ ### 2. $$\lim_{x \to 1^+} f(x)$$: For $$x > 1$$, the function is $$f(x) = 3x^2 + 2ax - 1$$. Evaluating this limit as $$x \to 1$$: $$\lim_{x \to 1^+} f(x) = 3(1)^2 + 2a(1) - 1 = 3 + 2a - 1 = 2a + 2.$$ ### 3. Continuity at $$x = 1$$: We are given that $$f(x)$$ is continuous at $$x = 1$$, so we require: $$\lim_{x \to 1^-} f(x) = f(1) = \lim_{x \to 1^+} f(x).$$ This gives us two conditions: - $$3b - 2 = 4a$$ (from $$\lim_{x \to 1^-} = f(1)$$). - $$2a + 2 = 4a$$ (from $$\lim_{x \to 1^+} = f(1)$$). ### Solving the system: From the second equation: $$2a + 2 = 4a \implies 2 = 2a \implies a = 1.$$ Substitute $$a = 1$$ into the first equation: $$3b - 2 = 4(1) \implies 3b - 2 = 4 \implies 3b = 6 \implies b = 2.$$ Thus, the solution is: $$a = 1, \quad b = 2.$$ Finally, the expression $$2a + 3b$$ can now be computed: $$2a + 3b = 2(1) + 3(2) = 2 + 6 = 8.$$ Would you like any further clarifications or details? Here are 5 related questions: 1. What is the general method to check for the continuity of piecewise functions? 2. How does L'Hopital's rule relate to solving limits involving trigonometric functions? 3. Can you give an example of a function that is not continuous and explain why? 4. What are the key conditions needed for differentiability in addition to continuity? 5. How would you determine if this function is differentiable at $$x = 1$$? **Tip:** Whenever you're solving for continuity in piecewise functions, carefully evaluate the one-sided limits and ensure they match the function value at the point in question.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Continuity of functions
Limits
Piecewise functions
Trigonometric limits
Formulas
Continuity condition: lim(x -> c^-) f(x) = f(c) = lim(x -> c^+) f(x)
Limit of trigonometric function: lim(x -> 0) (sin(x)/x) = 1
Theorems
Limit and continuity theorem
Squeeze theorem (for trigonometric limits)
Suitable Grade Level
Undergraduate level (Calculus I or II)