Calculus I, Fall 2015 (http://www.math.nthu.edu.tw/˜ wangwc/)

Brief answer to selected problems in HW07

1. Section 3.11: Problem 64.

(a)b_{0} = f(a), b_{1} = f′(a), and b_{2} = \frac{f′′(a)}{2} (b)Quadratic approximation of \frac{1}{1−x} at x = 0 is 1 + x + x^{2} (d)Quadratic approximation of \frac{1}{x} at x = 1 is 1 − (x − 1) + (x − 1)^{2} (e)Quadratic approximation of \sqrt{1 + x} at x = 0 is 1 + \frac{1}{2}x − \frac{1}{8}x^{2} Quadratic approximations are better than linear approximations near the base point x_{0} in general.

2. Error estimate for the linear approximation of section 3.11, problem 17 (b): For ^{3} \sqrt{1.009}, let

f(x) = (1 + x)^{\frac{1}{3}} ⟹ f′′(x) = −\frac{2}{9}(1 + x)^{− \frac{5}{3}}

Then ^{3} \sqrt{1.009} = f(x), x = 0.009. Hence we can use linear approximation near x_{0} = 0. By the error formula, we can get error bound by following process.

|f(0.009) − L(0.009; 0)| ≤ \frac{1}{2}( max_{0≤c≤0.009} |f′′(c)|)(0.009 − 0)^{2}
= \frac{1}{2} × \frac{2}{9} × 0.009^{2} = 9 × 10^{−6}

3. Chap 3, additional and advanced problem.

Problem 21: By definition, find the limit

lim_{x→x_{0}} \frac{f(x)g(x) − f(x_{0})g(x_{0})}{x − x_{0}}.

Use the fact that f(x_{0}) = 0 to replace f(x_{0})g(x_{0}) by f(x_{0})g(x). Then take the limit and conclude that answer = f′(x_{0})g(x_{0}). Problem 23:

(1) Find h′(x) for x ≠ 0 from straight forward calculation.

(2) Find h′(0) from definition h′(0) = lim_{x→0} \frac{h(x)−h(0)}{x}.

(3) Check whether or not lim_{x→0} h′(x) = h′(0). The procedure for k(x) = xh(x) is similar.