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Use linear approximation, i.e. the tangent line, to approximate \( \frac{1}{1.003} \) as follows: Let \( f(x)=\frac{1}{x} \) and find the equation of the tangent line to \( f(x) \) at a "nice" point near 1.003. Then use this to approximate \( \frac{1}{1.003} \) .

Ask by Nguyen Clark. in the United States
Mar 13,2025

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To approximate \( \frac{1}{1.003} \) using linear approximation: 1. Let \( f(x) = \frac{1}{x} \) and choose \( a = 1 \) as the point of approximation. 2. Compute \( f(1) = 1 \) and \( f'(x) = -\frac{1}{x^2} \), so \( f'(1) = -1 \). 3. The tangent line equation is \( L(x) = 2 - x \). 4. Substitute \( x = 1.003 \) into \( L(x) \) to get \( L(1.003) = 0.997 \). Therefore, \( \frac{1}{1.003} \approx 0.997 \).

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The Deep Dive

To approximate \( \frac{1}{1.003} \) using linear approximation, we first choose a nice point close to \( 1.003 \). The number \( x = 1 \) is a good choice because it makes calculations straightforward. Next, we compute the function \( f(x) = \frac{1}{x} \) and its derivative: 1. **Calculate \( f(1) \)**: \[ f(1) = \frac{1}{1} = 1 \] 2. **Find the derivative \( f'(x) \)**: \[ f'(x) = -\frac{1}{x^2} \] Therefore, at \( x = 1 \): \[ f'(1) = -\frac{1}{1^2} = -1 \] 3. **Form the equation of the tangent line at \( x = 1 \)**: The tangent line equation at a point \( (a, f(a)) \) is given by: \[ y - f(a) = f'(a)(x - a) \] Substituting in our values: \[ y - 1 = -1(x - 1) \] Simplifying, we get: \[ y = -x + 2 \] 4. **Use the tangent line to approximate \( f(1.003) \)**: Now we substitute \( x = 1.003 \) into the tangent line equation: \[ y \approx -1.003 + 2 = 0.997 \] Thus, using linear approximation, we find that \( \frac{1}{1.003} \approx 0.997 \).

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