# How to find image, preimage, domain and range of function? [closed]

Given:

(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 LahtonenNov 13 '18 at 4:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GEdgar, ArsenBerk, Chris Custer, Lee David Chung Lin, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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). – GEdgar Nov 12 '18 at 21:53

### Image

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 $$\{y \in \mathbb{R} | y > 0\}$$.

### Preimage

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 $$f^{-1}(A) = \{x \in mathbb{R}\setminus\{0\} | \frac{1}{x^2} \in A\} = \{\frac{1}{\sqrt{a}},\frac{-1}{\sqrt{a}} | a \in A\}$$.

### 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, "range" and "image" are synonyms.