## Calculus 8th Edition

a) 0 b) $\dfrac{e^2-1}{e^2+1}$
Always remember $\tanh = \dfrac{\sinh x}{\cosh x} =\dfrac{\dfrac{e^x-e^{-x}}{2}}{\dfrac{e^x+e^{-x}}{2}} =\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$ a) solve $\tanh 0$ $=\dfrac{e^0-e^{-0}}{e^0+e^{-0}}$ or ,$=\frac{1-1}{1+1}$ or, $=0$ b) solve $\tanh1$ $=\dfrac{e^1-e^{-1}}{e^1+e^{-1}}$ or, $=\dfrac{e-\frac{1}{e}}{e+\frac{1}{e}}$ or, $=\frac{e^2-1}{e^2+1}$