Linear approximation, question 13

Topic: Linear approximation
Difficulty: Capstone
Suggested time: 10 min
Marks: 5
Theory: UBC Math 100 §9.2 CLP-1 §3.4Key results

Let $f(x) = \log\!\left(1 + x^{2}\right)$ . Use the linear approximation of $f$ at $x = 1$ to estimate $f(1.1)$ and $f(0.9)$ . Then decide, for each estimate, whether it is an overestimate or an underestimate, and justify your answer.

Show your working and your reasoning, then compare against the model solution.

Your answer

Theory UBC Math 100 §9.2 CLP-1 §3.4Key results

Worked solution

Strategy

Build the tangent line at 1, where f and f′ are exact, and step 0.1 to each side. Then check the concavity on each step: 1 is an inflection point of this curve, so the two estimates err in opposite directions.

Working

$f(1) = \log 2, \qquad f'(x) = \frac{2x}{1 + x^{2}}, \qquad f'(1) = 1$
Chain rule: the derivative of log u is u′/u.
$L(x) = \log 2 + (x - 1): \qquad f(1.1) \approx \log 2 + \tfrac{1}{10}, \qquad f(0.9) \approx \log 2 - \tfrac{1}{10}$
Steps of +0.1 and −0.1 from the base point.
$f''(x) = \frac{2(1 + x^{2}) - 2x\cdot 2x}{(1 + x^{2})^{2}} = \frac{2(1 - x^{2})}{(1 + x^{2})^{2}}$
Quotient rule on 2x/(1 + x²); the denominator is positive, so f″ has the sign of 1 − x².
$1 < x \leq 1.1:\ f'' < 0 \quad\Longrightarrow\quad \log(2.21) < \log 2 + \tfrac{1}{10}$
Concave down, so the graph lies below its tangent line and this estimate is an overestimate.
$0.9 \leq x < 1:\ f'' > 0 \quad\Longrightarrow\quad \log(1.81) > \log 2 - \tfrac{1}{10}$
Concave up, so the graph lies above its tangent line and this estimate is an underestimate.

Answer

f(1.1) \approx \log 2 + \frac{1}{10}\ (\text{over}), \qquad f(0.9) \approx \log 2 - \frac{1}{10}\ (\text{under})

What earns the marks

Finds f(1) = log 2 and f′(1) = 1, so L(x) = log 2 + (x − 1).
Estimates f(1.1) ≈ log 2 + 1/10 and f(0.9) ≈ log 2 − 1/10.
Finds f″(x) = 2(1 − x²)/(1 + x²)² and notes that f″(1) = 0, so the second derivative at the base point decides nothing.
Uses the sign of f″ on each side of 1: negative for x > 1, so log 2 + 1/10 is an overestimate; positive for −1 < x < 1, so log 2 − 1/10 is an underestimate.

Check

log 2.21 ≈ 0.79299 sits just below log 2 + 0.1 ≈ 0.79315, and log 1.81 ≈ 0.59333 just above log 2 − 0.1 ≈ 0.59315.

Where this goes wrong

Deciding both from f″(1). It is 0 because 1 is an inflection point, so on its own it says nothing: the tangent line crosses the graph there, and the sign of f″ on each side of 1 decides.