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:**
Find the linear approximation to \( f(x) = \frac{1}{\sqrt{x}} \) at \( a = 4 \).
**Solution Overview:**
To find the linear approximation of a function at a given point, we use the formula for the tangent line approximation:
\[ L(x) = f(a) + f'(a)(x - a) \]
1. **Function:**
\[ f(x) = \frac{1}{\sqrt{x}} = x^{-0.5} \]
2. **Derivative:**
Using the power rule for differentiation:
\[ f'(x) = -0.5x^{-1.5} = -\frac{1}{2x^{1.5}} \]
3. **Evaluation at \( a = 4 \):**
- \( f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2} \)
- \( f'(4) = -\frac{1}{2(4)^{1.5}} = -\frac{1}{2 \times 8} = -\frac{1}{16} \)
4. **Linear Approximation:**
Substitute the above values in the linear approximation formula:
\[ L(x) = \frac{1}{2} + \left(-\frac{1}{16}\right)(x - 4) \]
Simplifying:
\[ L(x) = \frac{1}{2} - \frac{1}{16}(x - 4) \]
Thus, the linear approximation of \( f(x) = \frac{1}{\sqrt{x}} \) at \( x = 4 \) is:
\[ L(x) = \frac{1}{2} - \frac{1}{16}(x - 4) \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcdc8b149-2467-45b8-88fa-7bc1ca3ec305%2F6e714179-8ab9-456e-a330-0fc8fb3b0806%2Fxpy1ly_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the linear approximation to \( f(x) = \frac{1}{\sqrt{x}} \) at \( a = 4 \).
**Solution Overview:**
To find the linear approximation of a function at a given point, we use the formula for the tangent line approximation:
\[ L(x) = f(a) + f'(a)(x - a) \]
1. **Function:**
\[ f(x) = \frac{1}{\sqrt{x}} = x^{-0.5} \]
2. **Derivative:**
Using the power rule for differentiation:
\[ f'(x) = -0.5x^{-1.5} = -\frac{1}{2x^{1.5}} \]
3. **Evaluation at \( a = 4 \):**
- \( f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2} \)
- \( f'(4) = -\frac{1}{2(4)^{1.5}} = -\frac{1}{2 \times 8} = -\frac{1}{16} \)
4. **Linear Approximation:**
Substitute the above values in the linear approximation formula:
\[ L(x) = \frac{1}{2} + \left(-\frac{1}{16}\right)(x - 4) \]
Simplifying:
\[ L(x) = \frac{1}{2} - \frac{1}{16}(x - 4) \]
Thus, the linear approximation of \( f(x) = \frac{1}{\sqrt{x}} \) at \( x = 4 \) is:
\[ L(x) = \frac{1}{2} - \frac{1}{16}(x - 4) \]
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