Linear approximation
Problem 3.21 · easy
Use the linear approximation of \( \displaystyle f(x) = \frac{1}{x} \) at \( \displaystyle x = 2 \) to estimate \( \displaystyle f(\frac{99}{50}) \).
- The linearization at a is L(x) = f(a) + f'(a)(x − a).
- \[ \left. \frac{1}{x} \right|_{\substack{ x=2 }} = \frac{1}{2} \]f(a).✓ Proved
- \[ \left. \frac{d}{d x} \frac{1}{x} \right|_{\substack{ x=2 }} = - \frac{1}{4} \]f'(a).✓ Proved
- \[ 1 - \frac{x}{4} \]The linearization.✓ Proved
- \[ \frac{101}{200} \]Evaluate at x = \frac{99}{50}.✓ Proved
Answer \( L(\frac{99}{50}) = \frac{101}{200} \approx 0.505 \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the estimate is within Taylor's error bound of the true value |
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/linear_approximation, checked 2026-09-26 with SymPy 1.14.0.