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

(b) $f = \{((x_1, x_2), x_1−2 x_2)| x_1 > 0, x_2 > 0\}$

I need to find image, preimage, domain and range of these two and I don't know where to start.

If you guys could help me with at least one of the two functions, thanks!


closed as off-topic by GEdgar, ArsenBerk, Chris Custer, Lee David Chung Lin, Jyrki Lahtonen Nov 13 '18 at 4:16

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  • $\begingroup$ You don't know where to start? In most cases that means: start with the definitions. If your question is "put on hold", you could fix it up by adding the definition of "image" and showing what you get when you apply that definition to (a). $\endgroup$ – GEdgar Nov 12 '18 at 21:53

I'll do (a), and answer your four questions in order:


The image of $f$ is $\{f(x) | x \in \mathbb{R}\setminus\{0\}\}$. Note that for any $y > 0$, 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 $f(x) = \frac{1}{x^2} > 0$, so there's nothing else in the image. Thus, the image of $f$ is exactly $\{y \in \mathbb{R} | y > 0\}$.


This one's rather the odd one out: one doesn't have "a preimage of $f$", you have, for each subset $A \subseteq \mathop{\mathrm{Range}}(f)$, a preimage of $A$ under $f$, which is defined as $f^{-1}(A) = \{x \in \mathop{\mathrm{Domain}}(f) | f(x) \in A\}$. For your function, and any $A \subseteq \mathop{\mathrm{Range}}(f)$, we have $f^{-1}(A) = \{x \in mathbb{R}\setminus\{0\} | \frac{1}{x^2} \in A\} = \{\frac{1}{\sqrt{a}},\frac{-1}{\sqrt{a}} | a \in A\}$.


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].


Unless you're using some variant definitions that I'm not familiar with, "range" and "image" are synonyms.


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