How to find image, preimage, domain and range of function? Given:

(a) $f = \{(x, f(x))\mid x ∈ \mathbb R \setminus \{0\} , f(x) = 1/x^2\}$

(b) $f = \{((x_1, x_2),\, x_1−2 x_2)\mid x_1 &gt; 0, x_2 &gt; 0\}$

I need to find image, preimage, domain and range of these two and I don't know where to start.
I'll do (a), and answer your four questions in order:
Image
The image of $f$ is $\{f(x) \mid x \in \mathbb{R}\setminus\{0\}\}$. Note that for any $y &gt; 0$, we have $y = f\left(\frac{1}{\sqrt{y}}\right)$, so $y$ is in the image of $f$. Notice also that for every $x \in \mathbb{R}\setminus\{0\}$, we have $f(x) = \frac{1}{x^2} &gt; 0$, so there's nothing else in the image. Thus, the image of $f$ is exactly $\{y \in \mathbb{R} \mid y &gt; 0\}$.
Preimage
This one's rather the odd one out: one doesn't have &quot;a preimage of $f$&quot;, you have, for each subset $A \subseteq \mathop{\mathrm{Range}}(f)$, a preimage of $A$ under $f$, which is defined as $f^{-1}(A) = \{x \in \mathop{\mathrm{Domain}}(f) \mid f(x) \in A\}$. For your function, and any $A \subseteq \mathop{\mathrm{Range}}(f)$, we have $f^{-1}(A) = \{x \in \mathbb{R}\setminus\{0\} \mid \frac{1}{x^2} \in A\} = \{\frac{1}{\sqrt{a}},\frac{-1}{\sqrt{a}} \mid a \in A\}$.
Domain
This one's easy: it's explicitly given in the question. It's $\mathbb{R}\setminus\{0\}$. [Note that if it isn't given in the question, it's generally impossible to determine. This is the case, in particular, for (b). If you want to know why, ask away, and I'll have a rant].
Range
Unless you're using some variant definitions that I'm not familiar with, &quot;range&quot; and &quot;image&quot; are synonyms.
I'll do (a), and answer your four questions in order:
Image
The image of $f$ is $\{f(x) \mid x \in \mathbb{R}\setminus\{0\}\}$. Note that for any $y &gt; 0$, we have $y = f\left(\frac{1}{\sqrt{y}}\right)$, so $y$ is in the image of $f$. Notice also that for every $x \in \mathbb{R}\setminus\{0\}$, we have $f(x) = \frac{1}{x^2} &gt; 0$, so there's nothing else in the image. Thus, the image of $f$ is exactly $\{y \in \mathbb{R} \mid y &gt; 0\}$.
Preimage
This one's rather the odd one out: one doesn't have &quot;a preimage of $f$&quot;, you have, for each subset $A \subseteq \mathop{\mathrm{Range}}(f)$, a preimage of $A$ under $f$, which is defined as $f^{-1}(A) = \{x \in \mathop{\mathrm{Domain}}(f) \mid f(x) \in A\}$. For your function, and any $A \subseteq \mathop{\mathrm{Range}}(f)$, we have $f^{-1}(A) = \{x \in \mathbb{R}\setminus\{0\} \mid \frac{1}{x^2} \in A\} = \{\frac{1}{\sqrt{a}},\frac{-1}{\sqrt{a}} \mid a \in A\}$.
Domain
This one's easy: it's explicitly given in the question. It's $\mathbb{R}\setminus\{0\}$. [Note that if it isn't given in the question, it's generally impossible to determine. This is the case, in particular, for (b). If you want to know why, ask away, and I'll have a rant].
Range
Unless you're using some variant definitions that I'm not familiar with, &quot;range&quot; and &quot;image&quot; are synonyms.