Find the values of a and b that make f continuous everywhere. x² - 4 if x < 2 f(x) = x-2 ax² if 2 ≤ x < 3 if x ≥ 3 4x = a + b a b = 2-bx + 3

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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To determine the values of \(a\) and \(b\), the function \(f(x)\) must be continuous at the points where the pieces of the function connect, specifically at \(x = 2\) and \(x = 3\). This means the left-hand limit and right-hand limit must equal the function value at those points.

Given:
\[ f(x) = \begin{cases} 
\frac{x^2 - 4}{x - 2} & \text{if } x < 2 \\
ax^2 - bx + 3 & \text{if } 2 \leq x < 3 \\
4x - a + b & \text{if } x \geq 3 
\end{cases} \]

To ensure continuity at \(x = 2\):

Compute the left-hand limit:
\[\lim_{{x \to 2^-}} \frac{x^2 - 4}{x - 2} = \lim_{{x \to 2^-}} \frac{(x - 2)(x + 2)}{x - 2} = \lim_{{x \to 2^-}} (x + 2) = 4\]

Compute the right-hand limit at \(x = 2\):
\[\lim_{{x \to 2^+}} (ax^2 - bx + 3) = 4a - 2b + 3\]

Set these equal for continuity:
\[4 = 4a - 2b + 3\]

So,
\[1 = 4a - 2b \quad \text{(Equation 1)}\]

To ensure continuity at \(x = 3\):

Compute the left-hand limit:
\[\lim_{{x \to 3^-}} (ax^2 - bx + 3) = 9a - 3b + 3\]

Compute the right-hand limit:
\[\lim_{{x \to 3^+}} (4x - a + b) = 12 - a + b\]

Set these equal for continuity:
\[9a - 3b + 3 = 12 - a + b\]

So,
\[9a + a = 3b + b + 9\]
\[10a = 4b + 9\]

Divide by 2
Transcribed Image Text:To determine the values of \(a\) and \(b\), the function \(f(x)\) must be continuous at the points where the pieces of the function connect, specifically at \(x = 2\) and \(x = 3\). This means the left-hand limit and right-hand limit must equal the function value at those points. Given: \[ f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & \text{if } x < 2 \\ ax^2 - bx + 3 & \text{if } 2 \leq x < 3 \\ 4x - a + b & \text{if } x \geq 3 \end{cases} \] To ensure continuity at \(x = 2\): Compute the left-hand limit: \[\lim_{{x \to 2^-}} \frac{x^2 - 4}{x - 2} = \lim_{{x \to 2^-}} \frac{(x - 2)(x + 2)}{x - 2} = \lim_{{x \to 2^-}} (x + 2) = 4\] Compute the right-hand limit at \(x = 2\): \[\lim_{{x \to 2^+}} (ax^2 - bx + 3) = 4a - 2b + 3\] Set these equal for continuity: \[4 = 4a - 2b + 3\] So, \[1 = 4a - 2b \quad \text{(Equation 1)}\] To ensure continuity at \(x = 3\): Compute the left-hand limit: \[\lim_{{x \to 3^-}} (ax^2 - bx + 3) = 9a - 3b + 3\] Compute the right-hand limit: \[\lim_{{x \to 3^+}} (4x - a + b) = 12 - a + b\] Set these equal for continuity: \[9a - 3b + 3 = 12 - a + b\] So, \[9a + a = 3b + b + 9\] \[10a = 4b + 9\] Divide by 2
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