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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![**Problem Statement:**
Let \( f(x) = \begin{cases}
x^2 - a^2 x & \text{if } x < 2 \\
4 - 2x^2 & \text{if } x \geq 2
\end{cases} \).
Find all value(s) of \( a \) that make \( f \) continuous at 2. Justify your answer.
**Explanation:**
To ensure the function \( f(x) \) is continuous at \( x = 2 \), the following must be true:
1. The left-hand limit as \( x \) approaches 2 from the left, \( \lim_{{x \to 2^-}} f(x) \), must equal \( f(2) \).
2. The right-hand limit as \( x \) approaches 2 from the right, \( \lim_{{x \to 2^+}} f(x) \), must equal \( f(2) \).
3. Both limits must be equal, i.e., \( \lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^+}} f(x) \).
### Step-by-Step Solution:
**Step 1: Evaluate \( \lim_{{x \to 2^-}} f(x) \).**
Since the expression for \( x < 2 \) is \( x^2 - a^2 x \), substitute \( x = 2 \) into this expression:
\[ \lim_{{x \to 2^-}} f(x) = 2^2 - a^2 \times 2 = 4 - 2a^2 \]
**Step 2: Evaluate \( \lim_{{x \to 2^+}} f(x) \).**
For \( x \geq 2 \), the expression is \( 4 - 2x^2 \). Substitute \( x = 2 \):
\[ \lim_{{x \to 2^+}} f(x) = 4 - 2(2)^2 = 4 - 8 = -4 \]
**Step 3: Set the limits equal to ensure continuity at \( x = 2 \).**
\[ 4 - 2a^2 = -4 \]
Solve for \( a \):
\[ 4 - 2a^](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4e51140f-3bad-44fb-a755-4170f40982a7%2Fbe5ac5fd-a1de-4eeb-815d-e71f29d0287b%2Fgitywd_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Let \( f(x) = \begin{cases}
x^2 - a^2 x & \text{if } x < 2 \\
4 - 2x^2 & \text{if } x \geq 2
\end{cases} \).
Find all value(s) of \( a \) that make \( f \) continuous at 2. Justify your answer.
**Explanation:**
To ensure the function \( f(x) \) is continuous at \( x = 2 \), the following must be true:
1. The left-hand limit as \( x \) approaches 2 from the left, \( \lim_{{x \to 2^-}} f(x) \), must equal \( f(2) \).
2. The right-hand limit as \( x \) approaches 2 from the right, \( \lim_{{x \to 2^+}} f(x) \), must equal \( f(2) \).
3. Both limits must be equal, i.e., \( \lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^+}} f(x) \).
### Step-by-Step Solution:
**Step 1: Evaluate \( \lim_{{x \to 2^-}} f(x) \).**
Since the expression for \( x < 2 \) is \( x^2 - a^2 x \), substitute \( x = 2 \) into this expression:
\[ \lim_{{x \to 2^-}} f(x) = 2^2 - a^2 \times 2 = 4 - 2a^2 \]
**Step 2: Evaluate \( \lim_{{x \to 2^+}} f(x) \).**
For \( x \geq 2 \), the expression is \( 4 - 2x^2 \). Substitute \( x = 2 \):
\[ \lim_{{x \to 2^+}} f(x) = 4 - 2(2)^2 = 4 - 8 = -4 \]
**Step 3: Set the limits equal to ensure continuity at \( x = 2 \).**
\[ 4 - 2a^2 = -4 \]
Solve for \( a \):
\[ 4 - 2a^
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