Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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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: \[ L(x) = \frac{1}{2} - \frac{1}{16}(x - 4) \]
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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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