Find the linearization at x = a. f(x) = 1 a = 5 x3 (Express numbers in exact form. Use symbolic notation and fractions where needed.) L(x)
Find the linearization at x = a. f(x) = 1 a = 5 x3 (Express numbers in exact form. Use symbolic notation and fractions where needed.) L(x)
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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Question
![### Linearization at a Point
**Problem Statement:**
Find the linearization at \(x = a\).
Given:
\[
f(x) = \frac{1}{x^3}, \quad a = 5
\]
(Express numbers in exact form. Use symbolic notation and fractions where needed.)
\[
L(x) = \quad \_\_\_\_\_\_
\]
**Solution Guide:**
To find the linearization \(L(x)\) of the function \(f(x)\) at \(x = a\), we use the formula:
\[
L(x) = f(a) + f'(a)(x - a)
\]
Here’s the step-by-step solution process:
1. **Calculate \(f(a)\):**
\[
f(5) = \frac{1}{5^3} = \frac{1}{125}
\]
2. **Find the derivative \(f'(x)\):**
\[
f(x) = \frac{1}{x^3}
\]
\[
f'(x) = \frac{d}{dx} \left( x^{-3} \right) = -3x^{-4} = -\frac{3}{x^4}
\]
3. **Evaluate \(f'(a)\):**
\[
f'(5) = -\frac{3}{5^4} = -\frac{3}{625}
\]
4. **Substitute these values into the linearization formula:**
\[
L(x) = \frac{1}{125} - \frac{3}{625}(x - 5)
\]
5. **Simplify the expression:**
\[
L(x) = \frac{1}{125} - \frac{3}{625}(x - 5)
\]
Notice that \(\frac{3}{625} = \frac{3}{5 \cdot 125} = \frac{3}{5 \cdot 125} = \frac{1}{25 \cdot 25} = \frac{1}{25^2} = \frac{1}{625}\).
So the expression should be simplified and multiplied by sufficient constants to obtain:
\[
L(x) = \frac{1}{125} - \frac{3}{625}x + \frac{15}{625}
\]
Therefore, the final linearized function \( L(x) \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7369c9e1-cd27-40fa-b19e-4e793313e403%2F56f1f2ba-e06b-4ea1-baf8-304ba030d8d7%2Fhce1rur.png&w=3840&q=75)
Transcribed Image Text:### Linearization at a Point
**Problem Statement:**
Find the linearization at \(x = a\).
Given:
\[
f(x) = \frac{1}{x^3}, \quad a = 5
\]
(Express numbers in exact form. Use symbolic notation and fractions where needed.)
\[
L(x) = \quad \_\_\_\_\_\_
\]
**Solution Guide:**
To find the linearization \(L(x)\) of the function \(f(x)\) at \(x = a\), we use the formula:
\[
L(x) = f(a) + f'(a)(x - a)
\]
Here’s the step-by-step solution process:
1. **Calculate \(f(a)\):**
\[
f(5) = \frac{1}{5^3} = \frac{1}{125}
\]
2. **Find the derivative \(f'(x)\):**
\[
f(x) = \frac{1}{x^3}
\]
\[
f'(x) = \frac{d}{dx} \left( x^{-3} \right) = -3x^{-4} = -\frac{3}{x^4}
\]
3. **Evaluate \(f'(a)\):**
\[
f'(5) = -\frac{3}{5^4} = -\frac{3}{625}
\]
4. **Substitute these values into the linearization formula:**
\[
L(x) = \frac{1}{125} - \frac{3}{625}(x - 5)
\]
5. **Simplify the expression:**
\[
L(x) = \frac{1}{125} - \frac{3}{625}(x - 5)
\]
Notice that \(\frac{3}{625} = \frac{3}{5 \cdot 125} = \frac{3}{5 \cdot 125} = \frac{1}{25 \cdot 25} = \frac{1}{25^2} = \frac{1}{625}\).
So the expression should be simplified and multiplied by sufficient constants to obtain:
\[
L(x) = \frac{1}{125} - \frac{3}{625}x + \frac{15}{625}
\]
Therefore, the final linearized function \( L(x) \)
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