How do I show $y=\ \frac{1}{2}-\frac{1}{2^{x}+1\ }$ is odd?

Question

Of course I know that $f(-x)=-f(x)$, but I get confused as to how $$\frac{1}{2}-\frac{2^{x}}{2^{x}+1\ }=-\frac{1}{2}+\frac{1}{1+2^{x}}$$ In the steps of algebra.

Answer 1

Starting from where you left off:

$$\frac12 - \frac{2^x}{2^x+1} = \frac12-\left(\frac{2^x+1}{2^x+1}-\frac1{2^x+1}\right) = \frac12-1+\frac1{2^x+1}$$

Answer 2

So, I'd recommend turning the function into a proper fraction, so we get $$ f(x) = \dfrac{2^x-1}{2^{x+1}+2} $$

Therefore, $f(-x)$ is equal to $$ f(-x) = \dfrac{2^{-x}-1}{2^{-x+1}+2}$$

Multiply by $\dfrac{2^x}{2^x}$ and we get $$ f(-x) = \dfrac{1-2^x}{2^1+2\cdot 2^x} = \dfrac{1-2^x}{2^{x+1}+2} = -f(x)$$

Like we wanted.

Answer 3

There's really no trick:

$\frac{1}{2}-\frac{1}{2^{-x}+1\ }= \frac{1}{2^{x}+1\ }- \frac{1}{2} \Longleftrightarrow$

$1 = \frac{1}{2^{-x}+1\ }+ \frac{1}{2^{x}+1\ }$ and multiplying by the product of the denominators we get $\Longleftrightarrow$

$1= \frac{2^{x}+1 + 2^{-x}+1}{(2^{x}+1)(2^{-x}+1)}=\frac{2^{x}+1 + 2^{-x}+1}{2^{x}+1 + 2^{-x}+1}$. Since these steps are reversible the initial equality is true.