## Calculus: Early Transcendentals 8th Edition

a) $$0$$ b) $$\frac{e^2-1}{e^2+1}$$
To solve part a and b of this question, it is important to know that: $\tanh = \frac{\sinh x}{\cosh x} =\frac{\frac{e^x-e^{-x}}{2}}{\frac{e^x+e^{-x}}{2}} =\frac{e^x-e^{-x}}{e^x+e^{-x}}$ a) $\tanh 0$ $=\frac{e^0-e^{-0}}{e^0+e^{-0}}=\frac{1-1}{1+1}=\frac{0}{2}=0$ b) $\tanh1$ $=\frac{e^1-e^{-1}}{e^1+e^{-1}}=\frac{e-\frac{1}{e}}{e+\frac{1}{e}} =\frac{e^2-1}{e^2+1}$