A random sample of size $7$ is drawn from a distribution with p.d.f $$f_{\theta}(x)=\frac{1+x^2}{3\theta(1+\theta^2)}, -2\theta \le x \le \theta,\;x>0 \;\text{and otherwise zero}$$ and the observations are $12,-54,26,-2,24,17,-39$. What is the maximum likelihood estimate of $\theta$?
My Attempt:
Fix $x_1=12,x_2=-54,x_3=26,\cdots,x_7=-39$.
$L(\theta,x)=\prod_{i=1}^{7}\left(\frac{1+x_{i}^2}{3\theta(1+\theta^2)}\right)$. So, $$l(\theta,x)=\sum_{i=1}^{7}\log\left(\frac{1+x_{i}^2}{3\theta(1+\theta^2)}\right)$$ Now, $$\frac{\partial l}{\partial \theta}=\sum_{i=7}^{7}\left(\frac{3\theta(1+\theta^2)}{1+x_{i}^2}\right)\cdot \frac{-(3\theta(1+\theta^2)(1+x_{i}^2))}{(3\theta(1+\theta^2))^2}$$ Now, we just need to equate $\frac{\partial l}{\partial \theta}=0$ and find answer in terms of $x_i$
But after this, I am unable to find the MLE $\hat{\theta}$.
Please help me to solve this question. Thanks.
Hint:
$$f_\theta(x)=\frac{ 1+x^2}{3\theta(1+\theta^{2})} 1_{\{-2\theta \leq x\leq \theta\}}$$
$$f_\theta(x_1,\cdots,x_7)=\frac{\prod( 1+x_i^2)}{3^7\theta^7(1+\theta^{2})^{7}} \prod_{i=1}^{7} 1_{\{-2\theta \leq x_i \leq \theta\}}$$
The term $\prod_{i=1}^7 1_{\{-2\theta \leq x_i \leq \theta\}}$ is zero except if $\{-2\theta \leq \min (x_i)\leq \max (x_i) \leq \theta\}$. Define $M=\max (x_i)$ and $m=\min (x_i)$ so
$$f_\theta(x_1,\cdots,x_7)=\frac{\prod( 1+x_i^2)}{3^7\theta^7(1+\theta^{2})^{7}} 1_{\{M\leq \theta \}} 1_{\{-m/2 \leq \theta \}} ==\frac{\prod( 1+x_i^2)}{3^7\theta^7(1+\theta^{2})^{7}} 1_{\{\max(M,-m/2)\leq \theta \}} = \frac{\prod( 1+x_i^2)}{3^7\theta^7(1+\theta^{2})^{7}} 1_{\{27\leq \theta \}}.$$
Now maximize $$\frac{1}{3^7\theta^7(1+\theta^2)^7}$$ with condition
$$27\leq \theta$$