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Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). solution (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) we start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .

15 and 14.7 are 1 apart, so 15 – 14.7 must be 1.

Describe the end behavior of the function below. f(x)=(2/3)^x-2 a. as x increases, f(x) approaches infinity. b. as x decreases, f(x) approaches 2. c. as x increases, f(x) approaches -2. d. as x decreases, f(x) approaches negative infinity.