18.01 PROBLEM SET 3 – SOLUTIONS

2. Let x = 100 + h. Then

f(x) = 2 − \frac{50}{100 + h} − 2 · \frac{50^{2}}{(100 + h)^{2}} =

= 2 − \frac{50}{100 · (1 + \frac{h}{100})} − 2 · \frac{50^{2}}{100^{2} · (1 + \frac{h}{100})^{2}} =

= 2 − \frac{1}{2} · \frac{1}{1 + \frac{h}{100}} − 2 · \frac{1}{4} · \frac{1}{(1 + \frac{h}{100})^{2}} ≈

≈ 2 − \frac{1}{2} · (1 − \frac{h}{100}) − \frac{1}{2} · (1 − 2 · \frac{h}{100}) = 1 + \frac{3}{2} · \frac{h}{100}.

This gives f(101) ≈ 1 + 1.5 · 0.01 = 1.015.

(3 points)

3.

(1) lim_{θ→0} \frac{tan θ}{θ} = lim_{θ→0} \frac{sin θ}{θ} · lim_{θ→0} \frac{1}{cos θ} = 1 · 1 = 1.

(2) Let θ = tan^{−1} x. Then x = tan θ, and lim_{x→0} \frac{tan^{−1} x}{x} = lim_{θ→0} \frac{θ}{tan θ} = 1.

(3) f′(x) = \frac{x · \frac{1}{1 + x^{2}} − tan^{−1} x}{x^{2}} = \frac{\frac{1}{1 + x^{2}} − \frac{tan^{−1} x}{x}}{x} = \frac{\frac{1}{1 + x^{2}} − f(x)}{x}.

(4) f′′(x) = \frac{x · (−\frac{2x}{(1 + x^{2})^{2}} − f′(x)) − (\frac{1}{1 + x^{2}} − f(x))}{x^{2}} =

= −\frac{2}{(1 + x^{2})^{2}} − \frac{f′(x)}{x} − \frac{( \frac{\frac{1}{1 + x^{2}} − f(x)}{x})}{x} = −\frac{2}{(1 + x^{2})^{2}} − 2 · \frac{f′(x)}{x}.

(5) tan^{−1} x is odd, x is odd, so f(x) is even.

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(6) The tangent line of an even differentiable function is horizontal, because it has to be symmetric to the y-axis. Therefore its slope is zero.

(7) Since f′′(x) is continuous, using (4) we have

f′′(0) = lim_{x→0} f′′(x) = −2 − 2 · lim_{x→0} \frac{f′(x)}{x} = −2 − 2f′′(0).

Hence f′′(0) = −\frac{2}{3}.

(8) Using the standard quadratic approximation formula, (6) and (7) gives

\frac{tan^{−1} x}{x} ≈ f(0) + f′(0) · x + \frac{f′′(0)}{2} · x^{2} = 1 − \frac{x^{2}}{3} .

(9) By the double angle formula

\frac{\sqrt{3}}{3} = tan \frac{π}{6} = \frac{2 · tan \frac{π}{12}}{1 − tan^{2} \frac{π}{12}} .

This gives a quadratic equation for t = tan \frac{π}{12}

t^{2} + 2\sqrt{3} · t − 1 = 0.

t = tan \frac{π}{12} > 0, so the only solution is tan \frac{π}{12} = 2 − \sqrt{3}.

The half angle formula gives the same answer immediately.

(10) By (8) we have

\frac{\frac{π}{12}}{2 − \sqrt{3}} ≈ 1 − \frac{(2 − \sqrt{3})^{2}}{3} = \frac{3 − (4 − 4\sqrt{3} + 3)}{3} = \frac{4}{3}(\sqrt{3} − 1).

Hence

\frac{π}{12} ≈ \frac{4}{3}(2 − \sqrt{3})(\sqrt{3} − 1).

(12 points)