18.01 PROBLEM SET 2 – SOLUTIONS

2. First of all one can simplify the function. f(x) = \frac{x − 1}{(x − 2)^{2}(x + 1)}

domain: all real numbers except 2, −1. asymptotes: x = −1, x = 2, y = 0.

lim_{x→∞} f(x) = 0, lim_{x→−∞} f(x) = 0.

discontinuities: infinite discont. at x = −1, x = 2. zero: at x = 1.

(5 points)

3.

2 sinh x cosh x =2 · \frac{e^{x} − e^{−x}}{2} · \frac{e^{x} + e^{−x}}{2} = \frac{(e^{x})^{2} − (e^{−x})^{2}}{2} = sinh 2x

cosh^{2} x + sinh^{2} x = (\frac{e^{x} + e^{−x}}{2} )^{2} + (\frac{e^{x} − e^{−x}}{2} )^{2} =

= \frac{e^{2x} + 2 + e^{−2x} + e^{2x} − 2 + e^{−2x}}{4} = cosh 2x

cosh^{2} x − sinh^{2} x = (\frac{e^{x} + e^{−x}}{2} )^{2} − (\frac{e^{x} − e^{−x}}{2} )^{2} =

= \frac{e^{2x} + 2 + e^{−2x} − e^{2x} + 2 − e^{−2x}}{4} = 1

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2 18.01 PROBLEM SET 2 – SOLUTIONS

\frac{d}{dx} sinh x =\frac{d}{dx} \frac{e^{x} − e^{−x}}{2} = \frac{e^{x} − e^{−x} · (−1)}{2} = cosh x
\frac{d}{dx} cosh x =\frac{d}{dx} \frac{e^{x} + e^{−x}}{2} = \frac{e^{x} + e^{−x} · (−1)}{2} = sinh x.

The graphs of sinh x, cosh x, and sinh^{−1} x: Let y = sinh^{−1} x. Then x = sinh y, so

1 = \frac{d}{dx} sinh y = cosh y \frac{dy}{dx}.

From cosh^{2} y − sinh^{2} y = 1, we have cosh y = \sqrt{1 + sinh^{2} y} = \sqrt{1 + x^{2}} (note: cosh y > 0). Hence

\frac{d}{dx} sinh^{−1} x = \frac{dy}{dx} = \frac{1}{\sqrt{1 + x^{2}}} .

(10 points)