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\noindent \textbf{Problem Statement:}

Let \(f : \mathbb{R} \to \left(\infty, 2\right]\) be the function defined by
\(f(x) = 2-x^2\) for each \(x \in \mathbb{R}\). Show that \(f\) is surjective.


\noindent \textbf{Proof:}

To show that \(f\) is surjective, we need to show that it satisfies the property
\[
	\big(\forall y \in (-\infty, 2]\big) \, \left(\exists x \in \mathbb{R}\right) \, [f(x) = y]
\]

Suppose \(y\) is an arbitrary real number not greater than 2. Then, it remains to be shown that
there exists some real \(x\) such that \(f(x) = y\).

\textbf{Find an \(x\):} 
Let \(f(x) = y\). Then solving for \(x\), we obtain
\begin{align*}
	y &= 2-x^2	\\
	x^2 &= 2-y	\\
	x &= \sqrt{2-y}
\end{align*}
Thus, choose \(x = \sqrt{2-y}\). 

\textbf{Verify:} 
Next we must verify that this choice of \(x\) satisfies the domain and propositional function.

\begin{enumerate}
\item[\underline{Domain}:] {
	Since \(y \leq 2\), then \(2-y \geq 0\). This implies that \(x = \sqrt{2-y} \in \mathbb{R}\) as required.
	(Strictly speaking, it will be nonnegative; but that is irrelevant to this problem.)
}

\item[\underline{Proposition}:] {
	The choice of \(x\) must satisfy the propositional function \(f(x) = y\). Substitution yields
	\begin{align*}
		f(x) &= f \left(\sqrt{2-y}\right)	\\
			&= 2-\left(\sqrt{2-y}\right)^2	\\
			&= 2-(2-y)	\\
			&= 2-2+y	\\
			&= y
	\end{align*}
	Thus, \(f(x) = y\) as required.
}
\end{enumerate}
This proves the existential statement to be true. Furthermore, since our choice of \(y\) was arbitrary,
the original universal statement is true. This shows that the required property is satisfied; and this, in turn,
implies that the function \(f\) is surjective, as was to be proved.
\(\Box\)
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