Answer
a) $$0$$
b) $$\frac{e^2-1}{e^2+1}$$
Work Step by Step
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) $\tanh1$
$=\frac{e^1-e^{-1}}{e^1+e^{-1}}=\frac{e-\frac{1}{e}}{e+\frac{1}{e}}
=\frac{e^2-1}{e^2+1}$
