# Expressing hyperbolic functions in terms of $e$.

Express $\tanh(-3)$ in terms of $e$, where $\tanh$ is the hyperbolic tangent.

This is what I did:

\begin{align} \tanh(-x)&=\dfrac{e^{-2x}-1}{e^{-2x}+1}\\\\\\ \tanh(-3)&=\dfrac{e^{-2\times-3}-1}{e^{-2\times-3}+1}\\\\\\ \tanh(-3)&=\dfrac{e^6-1}{e^6+1} \end{align}

However, this is wrong, as the actual solution is:

$$\tanh(-3)=-\dfrac{e^3-1}{e^3+1}$$

1. What have I done that is unacceptable, hence making my solution wrong?

2. How is the actual solution obtained? (Full explanation would be helpful)

• You have too many minus signs! – Andrew Maurer Jan 19 '13 at 18:59
• I don't understand? – ODP Jan 19 '13 at 19:18
• You plug in $-3$ into the equation of $\tanh(-x)$, which gives you $\tanh(--3) = \tanh(3)$. – Andrew Maurer Jan 19 '13 at 20:05
• You shouldn't have the $2x$s in the second line. When you substitute in $-3$ for $x$, the $x$'s go away. – Ross Millikan Jan 19 '13 at 20:15
• That's not an $x$, that is a multiplication sign $\times$. – GEdgar Jan 19 '13 at 21:01

Using the definition $$\tanh(x) = \frac{e^{2x}-1}{e^{2x}+1}$$So we plug in $-3$ wherever we see an $x$ to get that $$\tanh(-3) = \frac{e^{2 \cdot-3}-1}{e^{2\cdot-3}+1}=\frac{e^{-6}-1}{e^{-6}+1}$$So we multiply by $\frac{e^6}{e^6}$ to get $$= \frac{1-e^6}{1 + e^6}$$So other than a little minus sign error, I think you're correct!
• Okay, but the definition I used for $tanh(-x)$ is the correct definition, in which case, surely I haven't done I minus sign error? Let me know where I'm going wrong with this statement please thanks for the answer as well – ODP Jan 20 '13 at 22:47
Your first attept is really right. In fact, you got $$\tanh(-3)=-\frac{e^3-e^{-3}}{e^3+e^{-3}}=-\frac{e^3-\frac{1}{e^{3}}}{e^3+\frac{1}{e^{-3}}}=-\frac{e^6-1}{e^6+1}$$ Knowing that $\tanh(x)$is an odd function also, the actual solution you pointed doesn't seem right result.
• numerator should be $e^{-3} - e^3$. – Andrew Maurer Jan 20 '13 at 7:31