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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fabc97ca2-5307-4b01-b84c-b32d2ca85d88%2F6a1ab363-37a3-4631-a0fd-068359456bba%2Fhf49ymk_processed.png&w=3840&q=75)
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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