se symbolic notation and fractions where neede
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Find the linearization of \( y = (1 + 7x)^{-1/2} \) at \( x = 5 \).
(Use symbolic notation and fractions where needed.)
\[ L(x) = \]
**Solution Explanation:**
To find the linearization of the function \( y = (1 + 7x)^{-1/2} \) at \( x = 5 \), follow these steps:
1. **Identify the function and point of linearization**:
\[
f(x) = (1 + 7x)^{-1/2}
\]
at \( x_0 = 5 \).
2. **Calculate \( f(x_0) \)**:
Substitute \( x = 5 \) into \( f(x) \):
\[
f(5) = (1 + 7 \cdot 5)^{-1/2} = (1 + 35)^{-1/2} = 36^{-1/2} = \frac{1}{6}
\]
3. **Determine \( f'(x) \)**:
Calculate the derivative \( f'(x) \). Using the chain rule:
\[
f(x) = (1 + 7x)^{-1/2} \implies f'(x) = -\frac{1}{2} (1 + 7x)^{-3/2} \cdot 7 = -\frac{7}{2} (1 + 7x)^{-3/2}
\]
Substitute \( x = 5 \) into \( f'(x) \):
\[
f'(5) = -\frac{7}{2} (1 + 35)^{-3/2} = -\frac{7}{2} \cdot 36^{-3/2}
\]
Since \( 36^{3/2} = (36^{1/2})^3 = 6^3 = 216 \):
\[
f'(5) = -\frac{7}{2} \cdot \frac{1}{216} = -\frac{7}{432}
\]
4. **Form the linearization**:
The linear approximation or linearization \( L(x) \) around \( x = x_0 \) is given by](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7369c9e1-cd27-40fa-b19e-4e793313e403%2F4b79aa8b-c870-4f70-970f-201a8c4ffcc9%2F4dk4r7q.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the linearization of \( y = (1 + 7x)^{-1/2} \) at \( x = 5 \).
(Use symbolic notation and fractions where needed.)
\[ L(x) = \]
**Solution Explanation:**
To find the linearization of the function \( y = (1 + 7x)^{-1/2} \) at \( x = 5 \), follow these steps:
1. **Identify the function and point of linearization**:
\[
f(x) = (1 + 7x)^{-1/2}
\]
at \( x_0 = 5 \).
2. **Calculate \( f(x_0) \)**:
Substitute \( x = 5 \) into \( f(x) \):
\[
f(5) = (1 + 7 \cdot 5)^{-1/2} = (1 + 35)^{-1/2} = 36^{-1/2} = \frac{1}{6}
\]
3. **Determine \( f'(x) \)**:
Calculate the derivative \( f'(x) \). Using the chain rule:
\[
f(x) = (1 + 7x)^{-1/2} \implies f'(x) = -\frac{1}{2} (1 + 7x)^{-3/2} \cdot 7 = -\frac{7}{2} (1 + 7x)^{-3/2}
\]
Substitute \( x = 5 \) into \( f'(x) \):
\[
f'(5) = -\frac{7}{2} (1 + 35)^{-3/2} = -\frac{7}{2} \cdot 36^{-3/2}
\]
Since \( 36^{3/2} = (36^{1/2})^3 = 6^3 = 216 \):
\[
f'(5) = -\frac{7}{2} \cdot \frac{1}{216} = -\frac{7}{432}
\]
4. **Form the linearization**:
The linear approximation or linearization \( L(x) \) around \( x = x_0 \) is given by
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