Answers: 3
Mathematics, 21.06.2019 19:50, Roshaan8039
Prove (a) cosh2(x) β sinh2(x) = 1 and (b) 1 β tanh 2(x) = sech 2(x). solution (a) cosh2(x) β sinh2(x) = ex + eβx 2 2 β 2 = e2x + 2 + eβ2x 4 β = 4 = . (b) we start with the identity proved in part (a): cosh2(x) β sinh2(x) = 1. if we divide both sides by cosh2(x), we get 1 β sinh2(x) cosh2(x) = 1 or 1 β tanh 2(x) = .
Answers: 3
Mathematics, 21.06.2019 20:30, maxy7347go
Does the function satisfy the hypotheses of the mean value theorem on the given interval? f(x) = 4x^2 + 3x + 4, [β1, 1] no, f is continuous on [β1, 1] but not differentiable on (β1, 1). no, f is not continuous on [β1, 1]. yes, f is continuous on [β1, 1] and differentiable on (β1, 1) since polynomials are continuous and differentiable on . there is not enough information to verify if this function satisfies the mean value theorem. yes, it does not matter if f is continuous or differentiable; every function satisfies the mean value theorem.
Answers: 1
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