solving inequality.

**Vim** · November 17th 2012, 03:42 AM

hi everyone. first of all i have to apologize for my poor English.
I've got an inequality which I know is true but don't know how to solve,can anyone help?
$2^{a}-2^{-a} \neq 2^{b}-2^{-b}$

**veileen** · November 17th 2012, 05:52 AM

That isn't an inequality.

Anyway, $2^a-2^{-a} = 2^b-2^{-b} \Leftrightarrow a=b$ .

I.

Let's consider the functions $f:\mathbb{R} \rightarrow \mathbb{R}, f(x)=2^x-2^{-x}$ , $f_1, f_2:\mathbb{R}\rightarrow \mathbb{R}, f_1(x)=2^x, f_2(x)=2^{-x}=\left ( \frac{1}{2} \right )^x$ .
$f_1$ is strictly increasing;
$f_2$ is strictly decreasing, so $-f_2$ is strictly increasing;

$f=f_1+f_2$ is strictly increasing (being a sum of two strictly increasing functions) $\Rightarrow$ f is injective, and therefore $f(a)=f(b) \Leftrightarrow a=b$ .

II.

$f:\mathbb{R} \rightarrow \mathbb{R}, f(x)=2^x-2^{-x}$

$f'(x)=(2^x-2^{-x})'=2^x\ln 2+2^{-x}\ln 2=\ln2(2^x+2^{-x})> 0$ , for any $x\in \mathbb{R} \Rightarrow$ f is injective, and therefore $f(a)=f(b) \Leftrightarrow a=b$ .

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