# I Pre Image

1. Jan 12, 2017

### mikeyBoy83

I'm trying to understand image and pre image better but I am having a hard time finding good examples.

So here is one I did come across, let's say $f:\mathbb{R} →\mathbb{R}$ defined by $f(x)=x^{2}$. Suppose also that $D = [-1,2]$ where $D\subset \mathbb{R}$. If I'm looking for $f^{-1}(D)$ then I can only use $D\setminus [-1,0)$ in which case $f^{-1}(D)=[-\sqrt{2},\sqrt{2}]$correct?

Last edited: Jan 12, 2017
2. Jan 12, 2017

### Lucas SV

Not correct. Rather, $f^{-1}(D) = [-\sqrt{2}, \sqrt{2}]$, since $f^{-1}(D)$ is defined as the set of all $x$ such that $f(x) \in D$. Also you can use $D$ and the inverse image is always well defined, even if $D$ is not completely contained in the range $f(\mathbb{R})$. In fact, $f^{-1}((-\infty, 2]) = [-\sqrt{2}, \sqrt{2}]$.

P.S. You really want double hashtags to begin and end a latex formula, rather than code

3. Jan 12, 2017

### mikeyBoy83

Can you explain why $f^{-1}(D) = [0,\sqrt{2}]$ ? Why would we not include $[-\sqrt{2},0]$ in our pre-image? Explain please.

4. Jan 12, 2017

### Lucas SV

5. Jan 12, 2017

### mikeyBoy83

Okay, so what if we wanted to find $f^{-1}(F)$ using the same function with $F=[-4,-1]$, in that case we would have $∅$.

6. Jan 12, 2017

### Lucas SV

Yes. Unless of course you decide to change domains to complex numbers.

7. Jan 12, 2017

Of course :)