= Solution
For observations $x=(x_1,\ldots,x_n)$, the <likelihood function> is
$$
L(\theta;x)=\prod_{i=1}^nf(x_i;\theta).
$$
A <maximum-likelihood estimator> is any measurable choice
$$
\boxed{\widehat\theta_{\mathrm{MLE}}(x)\in\operatorname*{argmax}_{\theta\in\Theta}L(\theta;x).}
$$
For one observation in a regular scalar model, the <score function> is
$$
S_\theta(X)=\frac{\partial}{\partial\theta}\log f(X;\theta),
$$
and the <Fisher information> is
$$
\boxed{I(\theta)=\mathbb E_\theta[S_\theta(X)^2]
=-\mathbb E_\theta\left[\frac{\partial^2}{\partial\theta^2}\log f(X;\theta)\right],}
$$
where the second identity requires the usual differentiation-under-the-integral regularity. For $n$ independent observations, <Tensorization of Fisher information> gives $I_n(\theta)=nI(\theta)$.
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