Exercise 1.4.33

Question

If you graph the function

$$
f(x)=\frac{1-e^{1 / x}}{1+e^{1 / x}}
$$

you'll see that $f$ appears to be an odd function. Prove it.

Khagan Gulusoy · Accepted Answer

\begin{proof}
To prove that the function $f$ is odd, we have to demonstrate that

$$
\forall x \in X: f(-x)=-f(x) 	ext {, }
$$

where $X$ is the domain of the function $f$. Therefore, first of all, we should find $f(-x)$ and $-f(x)$ :

$$
\begin{aligned}
& f(-x)=\frac{1-e^{1 / x}}{1+e^{1 / x}}=\frac{1-e^{-1 / x}}{1+e^{-1 / x}} ; \
& -f(x)=-\frac{1-e^{1 / x}}{1+e^{1 / x}}=\frac{e^{1 / x}-1}{1+e^{2 / x}} .
\end{aligned}
$$

Therefore, the statement we have to prove is equivalent to the following statement:

$$
\begin{aligned}
& \forall x \in X: \frac{1-e^{-1 / x}}{1+e^{-1 / x}}=\frac{e^{1 / x}-1}{1+e^{2 / x}} \
& \iff\left(1-e^{-1 / x}ight)\left(1+e^{1 / x}ight)=\left(e^{1 / x}-1ight)\left(1+e^{-1 / x}ight) \
& \iff 1-e^{-1 / x}+e^{1 / x}-e^{-1 / x} \cdot e^{1 / x}=e^{1 / x}-1+e^{1 / x} \cdot e^{-1 / x}-e^{-1 / x} 	ext {. } \
& \iff 1-e^{-1 / x}+e^{1 / x}-e^{-1 / x+1 / x}=e^{1 / x}-1+e^{1 / x-1 / x}-e^{-1 / x} . \
& \iff 1-e^{-1 / x}+e^{1 / x}-e^{0}=e^{1 / x}-1+e^{0}-e^{-1 / x} . \
& \iff 1-e^{-1 / x}+e^{1 / x}-1=e^{1 / x}-1+1-e^{-1 / x} . \
& \iff-e^{-1 / x}+e^{1 / x}=e^{1 / x}-e^{-1 / x} 	ext {. }
\end{aligned}
$$

The last statement we obtained is always true, as desired.
\end{proof}

Exercise 1.4.33

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Exercise 1.4.33

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