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Chapter 1 | Functions and Graphs
Let’s look at how to derive the first equation. The others follow similarly. Suppose y = sinh −1 x . Then, x = sinh y and, by the definition of the hyperbolic sine function, x = e y − e − y 2 . Therefore, e y −2 x − e − y =0. Multiplying this equation by e y , we obtain e 2 y −2 xe y −1=0. This can be solved like a quadratic equation, with the solution e y = 2 x ± 4 x 2 +4 2 = x ± x 2 +1. Since e y >0, the only solution is the one with the positive sign. Applying the natural logarithm to both sides of the equation, we conclude that y = ln ⎛ ⎝ x + x 2 +1 ⎞ ⎠ .
Example 1.41 Evaluating Inverse Hyperbolic Functions
Evaluate each of the following expressions.
sinh −1 (2) tanh −1 (1/4)
Solution sinh −1 (2) = ln ⎛
⎝ 2+ 2 2 +1 ⎞
⎠ = ln ⎛
⎝ 2+ 5 ⎞
⎠ ≈1.4436
⎛ ⎝ 1+1/4 1−1/4
⎞ ⎠ = 1 2 ln ⎛
⎞ ⎠ = 1 2 ln ⎛
⎞ ⎠ ≈0.2554
⎝ 5/4 3/4
⎝ 5 3
tanh −1 (1/4) = 1
2 ln
Evaluate tanh −1 (1/2).
1.35
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