Consider the following piecewise defined function ae*-b, x < 0 ax² +b, 0≤x≤2 a(x − 3)³ + b(x − 4) + 26, x ≥ 2 where a and b are real constants. For what values of a and b is f(x) continuous? f(x) = faez For no real values of a and b, f(x) is continuous. For any pair (a, b) where a = 2b, f(x) is continuous. For any pair (a, b) where 5a + 3b = Only for (a, b) = (2,4) Only for (a, b) = (4,2) 26, f(x) is continuous.
Consider the following piecewise defined function ae*-b, x < 0 ax² +b, 0≤x≤2 a(x − 3)³ + b(x − 4) + 26, x ≥ 2 where a and b are real constants. For what values of a and b is f(x) continuous? f(x) = faez For no real values of a and b, f(x) is continuous. For any pair (a, b) where a = 2b, f(x) is continuous. For any pair (a, b) where 5a + 3b = Only for (a, b) = (2,4) Only for (a, b) = (4,2) 26, f(x) is continuous.
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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![### Piecewise Function and Continuity
**Consider the following piecewise defined function:**
\[
f(x) = \begin{cases}
a e^x - b, & x < 0 \\
a x^2 + b, & 0 \leq x < 2 \\
a(x - 3)^3 + b(x - 4) + 26, & x \geq 2
\end{cases}
\]
**where \(a\) and \(b\) are real constants. For what values of \(a\) and \(b\) is \(f(x)\) continuous?**
**Options:**
1. \( \bigcirc \) For no real values of \(a\) and \(b\), \(f(x)\) is continuous.
2. \( \bigcirc \) For any pair \((a, b)\) where \(a = 2b\), \(f(x)\) is continuous.
3. \( \bigcirc \) For any pair \((a, b)\) where \(5a + 3b = 26\), \(f(x)\) is continuous.
4. \( \bigcirc \) Only for \((a, b) = (2, 4)\)
5. \( \bigcirc \) Only for \((a, b) = (4, 2)\)
### Explanation
In this piecewise function, \(f(x)\) is defined differently over three intervals:
1. **For \(x < 0\):** \(f(x) = a e^x - b\)
2. **For \(0 \leq x < 2\):** \(f(x) = a x^2 + b\)
3. **For \(x \geq 2\):** \(f(x) = a(x - 3)^3 + b(x - 4) + 26\)
In order for \(f(x)\) to be continuous at all points, the following must hold true:
- The function must match at the boundaries of the intervals.
For \(f(x)\) to be continuous at \(x = 0\):
\[
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)
\]
\[
ae^0 - b = a(0)^2 + b](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F49dd86f5-e3e4-495a-9b19-f3d039df6a7c%2F3f8832aa-7214-4a3b-9354-684ae5505abb%2Fzer3paa_processed.png&w=3840&q=75)
Transcribed Image Text:### Piecewise Function and Continuity
**Consider the following piecewise defined function:**
\[
f(x) = \begin{cases}
a e^x - b, & x < 0 \\
a x^2 + b, & 0 \leq x < 2 \\
a(x - 3)^3 + b(x - 4) + 26, & x \geq 2
\end{cases}
\]
**where \(a\) and \(b\) are real constants. For what values of \(a\) and \(b\) is \(f(x)\) continuous?**
**Options:**
1. \( \bigcirc \) For no real values of \(a\) and \(b\), \(f(x)\) is continuous.
2. \( \bigcirc \) For any pair \((a, b)\) where \(a = 2b\), \(f(x)\) is continuous.
3. \( \bigcirc \) For any pair \((a, b)\) where \(5a + 3b = 26\), \(f(x)\) is continuous.
4. \( \bigcirc \) Only for \((a, b) = (2, 4)\)
5. \( \bigcirc \) Only for \((a, b) = (4, 2)\)
### Explanation
In this piecewise function, \(f(x)\) is defined differently over three intervals:
1. **For \(x < 0\):** \(f(x) = a e^x - b\)
2. **For \(0 \leq x < 2\):** \(f(x) = a x^2 + b\)
3. **For \(x \geq 2\):** \(f(x) = a(x - 3)^3 + b(x - 4) + 26\)
In order for \(f(x)\) to be continuous at all points, the following must hold true:
- The function must match at the boundaries of the intervals.
For \(f(x)\) to be continuous at \(x = 0\):
\[
\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)
\]
\[
ae^0 - b = a(0)^2 + b
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