Solve the inequality $(1/2)^x-(1/2)^{-1-x}\ge1$ for real $x$

Question

I have to solve in $\Bbb{R}$ the following inequality :

$$ \left(\frac{1}{2}\right)^{x} - \left(\frac{1}{2}\right)^{-1 - x} \ge 1 \qquad(E) $$

So far I have :

For $x=0$ this inequality if not true. Writting $a^x=\exp(x\log(a))$ $(E)$ can be written $$ 2^{-x}-2^{1+x}\ge1 $$

I tried to factor but I do not get a friendly solution, and then I tried to study : $$ {\rm f}\left(x\right)=2^{-x}-2^{1+x} $$ I have ${\rm f}'\left(x\right)=2^{-x}(1+2^{1+2 x})\log(2)$ wich is stricly negative for all $x$, and so ${\rm f}$ is decreasing.

We have ${\rm f}\left(x\right) \rightarrow +\infty$ as $x$ tends to $-\infty$ and ${\rm f}\left(-1\right)=1$ then the inequality is correct only for $x \le 1$.

I think it's correct ( ? ) but do you think this can be solved without using the derivative ?.

Answer 1

Let us take up things when you reached $2^{-x}-2^{1+x}\ge 1$. Multiply through by $2^x$, and rearrange a bit. We get $2\cdot 2^{2x}+2^x-1\ge 0$. Rewrite as $(2^x+1)(2\cdot 2^x-1)\ge 0$.

Answer 2

Hint: transform the inequality to a quadratic inequality with the substitution $y = 2^x$, then $2^{-x} = \dfrac{1}{y}$