Calculating $P\{\max(-x_{(1)},x_{(n)})\leq x \}$ where $X_{i}$ are i.i.d. from uniform distribution

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Let $X_{i}$ be i.i.d. from uniform distribution $[-\theta,2\theta]$. I am interested in if the following is true:

$P\{\max(-x_{(1)},x_{(n)})\leq x \}= P\{-x_{(1)}\leq x , x_{(n)}\leq x \}\\ = P\{-x\leq x_{(1)} , x_{(n)}\leq x \}\\= P\{-x\leq x_{1} \leq x, ... ,-x\leq x_{(n)}\leq x \}\\= P\{-x\leq x_{1} \leq x \} ... P\{-x\leq x_{n} \leq x \}\\=(\int_{-x}^{x}\frac{1}{3\theta})^{n} $

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The last line is not quite correct since $x$ is not known. You will have to make cases:

$$F_{\max\{-x_{(1)},x_{(n)}\}}(x)=\begin{cases}\int_{-x}^x\frac{dy}{3\theta},&0< x\le\theta\\\int_{-\theta}^x\frac{dy}{3\theta},&\theta< x\le2\theta\\1,&2\theta< x\\0,&\text{otherwise}\end{cases}$$