Tutorial Video:Continuity -Example 2
Solving for Continuity in Piecewise Functions
The Problem
Find the values of a and b that make the function \(f(x)\) continuous on \((-\infty, \infty)\):
\[ f(x)=\begin{cases} \frac{x^3-8}{x-2} & \text{if } x < 2 \\ ax^3-bx+3 & \text{if } 2 \leq x < 3 \\ 2x-a+b & \text{if } x \geq 3 \end{cases} \]
Step 1: Analyzing the Junction at \(x = 2\)
For continuity at \(x = 2\), the left-hand limit (LHL) must equal the right-hand limit (RHL).
Left-hand Limit (LHL):
\[ \lim_{x \to 2^-} \frac{x^3-8}{x-2} \] Using L'Hôpital's Rule: \( \frac{d/dx(x^3-8)}{d/dx(x-2)} = \frac{3x^2}{1} \)
Evaluating at \(x=2\): \( 3(2)^2 = \mathbf{12} \)
\[ \lim_{x \to 2^-} \frac{x^3-8}{x-2} \] Using L'Hôpital's Rule: \( \frac{d/dx(x^3-8)}{d/dx(x-2)} = \frac{3x^2}{1} \)
Evaluating at \(x=2\): \( 3(2)^2 = \mathbf{12} \)
Right-hand Limit (RHL):
\[ \lim_{x \to 2^+} (ax^3 - bx + 3) = 8a - 2b + 3 \]
\[ \lim_{x \to 2^+} (ax^3 - bx + 3) = 8a - 2b + 3 \]
Equating LHL and RHL: \( 12 = 8a - 2b + 3 \implies \) \( 8a - 2b = 9 \) — (Eq. 1)
Step 2: Analyzing the Junction at \(x = 3\)
Next, we ensure the pieces meet at the second transition point, \(x = 3\).
LHL at 3: \( 27a - 3b + 3 \)
RHL at 3: \( 2(3) - a + b = 6 - a + b \)
RHL at 3: \( 2(3) - a + b = 6 - a + b \)
Equating these gives: \( 27a - 3b + 3 = 6 - a + b \)
Rearranging terms: \( 28a - 4b = 3 \) — (Eq. 2)
Step 3: Solving the System
Using substitution or elimination on the two equations:
- \( 8a - 2b = 9 \)
- \( 28a - 4b = 3 \)
Final Answer: \( a = -\frac{5}{4}, \quad b = -\frac{19}{2} \)
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