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\begin{parts}

\part The range of the hyperbolic tangent function $\tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}$ is $(-1,1)$. Use the fact that $e^{anything} > 0$ to explain why $2$ and $-3$ are not in the range of $\tanh(x)$.
% \tanh(x)=2 means e^x - e^{-x} = 2e^x + 2e^{-x} means -e^x = 3e^{-x} which is impossible since the LHS is positive and the RHS is negative
% \tanh(x) = -3 means e^x - e^{-x} = -3e^x - 3e^{-x} means 2e^x = -2e^{-x} which is impossible since the LHS is positive and the RHS is negative

\part Show that $\tanh(x)$ is always strictly increasing, and hence that it is one-to-one.

\part What are the domain and range of the inverse function $\tanh^{-1}(x)$?

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