Calculus I, Fall 2015

Quiz 3

Nov 12, 2015

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1. Find an approximate value of \sqrt[3]{1.009} and show that the error of the approximation is smaller than 10^{−5}.

Ans: (i) Let f(x) = (1 + x)^{\frac{1}{3}} and L(x, 0) = f(0) + f′(0)(x − 0) = 1 + \frac{1}{3}x(10 points) then \sqrt[3]{1.009} = f(0.009) ≈ L(0.009) = 1.003(5 points) (ii)

|f(0.009) − L(0.009)| ≤ \frac{1}{2} max_{0≤c≤0.009} |f′′(c)|(0.009 − 0)^{2} = \frac{81}{2} max_{0≤c≤0.009} | − \frac{2}{9}(1 + x)^{− \frac{5}{3}} | · 10^{−6}
= 9 · 10^{−6} ≤ 10^{−5}(10points)

2. Show that, if Rolle’s Theorem is true, then the Mean Value Theorem is true.

Ans: If f(x) is continuous on [a, b] and differentiable on (a, b), let g(x) = f(x) − [f(a) + \frac{f(b)−f(a)}{b−a}(x − a)].(10 points) Then g(x) is also continuous on [a, b] and differentiable on (a, b) with g(a) = g(b).(8 points) Hence by Rolle’s Theorem, ∃c ∈ (a, b) such that g′(c) = 0, which implies f′(c) = \frac{f(b)−f(a)}{b−a}.(7 points)

3. Let f(x) = x^{2/3}(3 − x). Find absolute max and min of f on [−3, 3]. Show all details.

Ans: f′(x) = \frac{2}{3}x^{− \frac{1}{3}} (3 − x) − x^{\frac{2}{3}} = \frac{6−5x}{3x^{\frac{1}{3}}}. (6 points) Critical points: x = 0, \frac{6}{5}. (4 points) f(0) = f(3) = 0 < f(\frac{6}{5}) = \frac{9}{5} · (\frac{6}{5})^{\frac{2}{3}} < f(−3) = 6 · 3^{\frac{2}{3}} . (9 points) Absolute maximum: f(−3) = 6 · 3^{\frac{2}{3}} . (3 points) Absolute minimum: f(0) = f(3) = 0. (3 points)

4. Graph y = \frac{1}{x^{2}−1}. Show all details.

Ans: Domain: x ≠ ±1. y′ = \frac{−2x}{(x^{2}−1)^{2}} , x ≠ ±1. (3 points)

y′′ = \frac{−2(x^{2}−1)^{2}−(−2x)·2(x^{2}−1)2x}{(x^{2}−1)^{4}} = \frac{6x^{2}+2}{(x^{2}−1)^{3}} , x ≠ ±1. (3 points)

Increasing interval: (−∞, −1), (−1, 0], Decreasing interval: [0, 1), (1, ∞). (6 points) Concave up: (−∞, −1), (1, ∞), Concave down: (−1, 1). (6 points) Critical points: x = 0. Local maximum: f(0) = −1. (1 point) Vertical asymptote: lim_{x→±1} y = ±∞ ⇒ x = ±1 are vertical asymptotes. Horizontal asymptote: lim_{x→±∞} y = 0 ⇒ y = 0 is a horizontal asymptote. Graph: (6 points)