To solve the problem, we need to handle inverse images of sets under the function $f(x) = x^2$ where $f: \mathbb{R} \to \mathbb{R}$.

1. Understanding the inverse image of a set:

The inverse image $f^{-1}(S)$ of a subset $S \subseteq \mathbb{R}$ is defined as: $f^{-1}(S) = \{a \in A \mid f(a) \in S\}.$ In this case, $f(x) = x^2$, so we're looking for all values of $x \in \mathbb{R}$ such that $x^2 \in S$.

Now, let's find the inverse images of the given sets:

2. Finding $f^{-1}(\{1\})$:

We are asked to find $f^{-1}(\{1\})$, meaning we need to find all $x \in \mathbb{R}$ such that: $f(x) = x^2 = 1.$ This gives two possible solutions: $x^2 = 1 \quad \Rightarrow \quad x = \pm 1.$ Thus, $f^{-1}(\{1\}) = \{-1, 1\}$.

3. Finding $f^{-1}(\{x \mid 0 < x < 1\})$:

Next, we need to find $f^{-1}(\{x \mid 0 < x < 1\})$, which means finding all $x \in \mathbb{R}$ such that: $0 < x^2 < 1.$ The solutions to this inequality are: $-1 < x < 1.$ Thus, $f^{-1}(\{x \mid 0 < x < 1\}) = (-1, 1)$.

4. Finding $f^{-1}(\{x \mid x > 4\})$:

Finally, we need to find $f^{-1}(\{x \mid x > 4\})$, which means finding all $x \in \mathbb{R}$ such that: $x^2 > 4.$ This inequality holds when: $x > 2 \quad \text{or} \quad x < -2.$ Thus, $f^{-1}(\{x \mid x > 4\}) = (-\infty, -2) \cup (2, \infty)$.

Final Answers:

1. $f^{-1}(\{1\}) = \{-1, 1\}$
2. $f^{-1}(\{x \mid 0 < x < 1\}) = (-1, 1)$
3. $f^{-1}(\{x \mid x > 4\}) = (-\infty, -2) \cup (2, \infty)$

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Do you want details or have any questions?

Here are 5 related questions to explore further:

1. What is the inverse image of a set for a different function, like $f(x) = x^3$?
2. How does the inverse image change when the function involves absolute values, like $f(x) = |x|$?
3. Can we generalize the process of finding inverse images for any polynomial function?
4. What happens to the inverse image if we restrict the function $f(x) = x^2$ to positive real numbers only?
5. How does the concept of inverse images relate to preimages in topology?

Tip: Always remember that the inverse image of a set might contain more than one value, even if the original set is a single point. This happens especially with non-injective functions like $f(x) = x^2$.