Likelihood ratio test for composite alternative hypothesis I must admit that my general understanding on how to create critical region to test hypothesis against composite alternative hypothesis is still shaky. Therefore please pardon if my question is too trivial.
Lets take one example from Likelihood ratio test to determine if average number of accidents has dropped?
Here $H_1: \lambda < 4.$
It is mentioned that general form of likelihood ratio test takes the form as
$$\Lambda(x)=\frac{\sup_{\lambda\in\Theta_{H_{0}}}L(\lambda|x)}{\sup_{\lambda\in\Theta}L(\lambda|x)} < c. $$
I think I understand why this is so, as in order to reject the null, I must have higher likelihood under the alternate hypothesis given a fixed sample. Therefore numerator must be smaller to the denominator.
After following some direct mathematics final critical region comes as $\sum x<c$
However it appears that in this entire derivation we did not take the direction of $H_1$ which is $H_1: \lambda < 4.$
How the critical region would change if I had $H_1: \lambda > 4$ or $H_1: \lambda \neq 4$?

Here $\Theta_0:=\{\lambda: \lambda =4\}$ and $\Theta:=\{\lambda:\lambda\leq 4\}.$
Now \begin{align}\frac{\mathrm d}{\mathrm d\lambda}\mathcal L(\lambda|\mathbf x) &=\frac{\mathrm d}{\mathrm d\lambda}\left[-n\ln \lambda-\frac{ n\bar { x}}{\lambda}\right]\\ &= -\frac n\lambda+\frac{n\bar{ x}}{\lambda^2}.\tag 1\label 1\end{align} From $\eqref 1,$ $$\frac{\mathrm d}{\mathrm d\lambda}\mathcal L(\lambda|\mathbf x) > 0\implies \lambda <\bar x .\tag 2\label 2$$
If ${\bar x}\in(0, 4],$ then from $\eqref 2,~\hat{\lambda}_{\text{MLE}}={\bar x}.$ If ${\bar x}\in(4, \infty),$ then $\hat{\lambda}_{\text{MLE}} =4.$
So, if ${\bar x}\in(4, \infty),$ $$\ell(\mathbf x) =1\tag{ 3.I}\label 3$$ and if ${\bar x}\in(0, 4],$ then $$\ell(\mathbf x) =\frac{4^{-n}\exp\left(-\frac{n\bar x}{4}\right)}{(1/\bar x)^n\exp(-n)}.\tag{3.II}$$
$\eqref 3$ is trivial.
Observe $$\frac{\mathrm d}{\mathrm d\bar x}\ell(\mathbf x) =k\times \left[n{\bar x}^{n-1}\exp\left(\frac{-n\bar x}{4}\right)\left(1-\frac{\bar x}4\right)\right].$$
So, when ${\bar x}\in(0, 4],$
$$\ell(\mathbf x) < \textrm{some constant}\iff \bar x< \textrm{some other constant}.$$
The test function then follows.
Similarly, one can proceed along the same approach for $\mathcal H_1: \lambda > 4.$
