9. Let f be the function from R to R defined by f (x) = x2. Finda) f −1({1}) b) f −1({x | 0 < x < 1}) c) f −1({x | x > 4})

1. Break Down the Problem

We need to find the inverse image (or preimage) of the function $f(x) = x^2$ for three different sets:

a) $f^{-1}(\{1\})$
b) $f^{-1}(\{x \ | \ 0 < x < 1\})$
c) $f^{-1}(\{x \ | \ x > 4\})$

2. Relevant Concepts

The inverse image $f^{-1}(A)$ for a function $f : \mathbb{R} \to \mathbb{R}$ is defined as:
$f^{-1}(A) = \{ x \in \mathbb{R} \ | \ f(x) \in A \}$

3. Analysis and Detail

a) Finding $f^{-1}(\{1\})$

We need to solve the equation:
$f(x) = 1 \implies x^2 = 1$ This gives us:
$x = 1 \quad \text{or} \quad x = -1$

Thus,
$f^{-1}(\{1\}) = \{1, -1\}$

b) Finding $f^{-1}(\{x \ | \ 0 < x < 1\})$

We need to solve the inequality:
$0 < f(x) < 1 \implies 0 < x^2 < 1$ Taking square roots (and noting the restriction on $x$):
$0 < |x| < 1 \implies -1 < x < 1 \quad \text{and} \quad x \neq 0$

So, the result is:
$f^{-1}(\{x \ | \ 0 < x < 1\}) = (-1, 0) \cup (0, 1)$

c) Finding $f^{-1}(\{x \ | \ x > 4\})$

We need to solve the inequality:
$f(x) > 4 \implies x^2 > 4$ Taking square roots gives us:
$|x| > 2 \implies x < -2 \quad \text{or} \quad x > 2$

Thus,
$f^{-1}(\{x \ | \ x > 4\}) = (-\infty, -2) \cup (2, \infty)$

4. Verify and Summarize

Let's summarize the findings:

a) $f^{-1}(\{1\}) = \{1, -1\}$
b) $f^{-1}(\{x \ | \ 0 < x < 1\}) = (-1, 0) \cup (0, 1)$
c) $f^{-1}(\{x \ | \ x > 4\}) = (-\infty, -2) \cup (2, \infty)$

Final Answer

a) $f^{-1}(\{1\}) = \{1, -1\}$
b) $f^{-1}(\{x \ | \ 0 < x < 1\}) = (-1, 0) \cup (0, 1)$
c) $f^{-1}(\{x \ | \ x > 4\}) = (-\infty, -2) \cup (2, \infty)$