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Let $f:\mathbb{R}\to\mathbb{R}$ be a function defined as follows:

$$f(x) = \begin{cases} x^2 : x>0\\ 1-x : x\leq0\\ \end{cases}$$

I have to calculate the pre-image of the interval $[-2,3)$

We know that the function's image are the positive real numbers, then we can simplify the interval this way:

$$f^{-1}([-2,3))=f^{-1}((0,3))\cup f^{-1}([-2,0])=f^{-1}((0,3))$$

Because negative images don't exist.

To calculate $f^{-1}((0,3))$ I did this:

For the positive part of the function, $f(x)=x^2$

$f^{-1}((0,3))=(x^2=0,x^2=3)=(0,\sqrt3)$

For the negative part of the function, $f(x)=1-x$

$f^{-1}((0,3))=(1-x=0,1-x=3)=(1,-2)$

$f(x)=1-x$ is defined only for negative numbers and 0, then:

$f^{-1}((0,3))=(-2,1)-(0,1)=(-2,0]$

Finally, the pre-image of the function should be the sum of the two intervals:

$f^{-1}([-2,3)) = (0,\sqrt3)\cup (-2,0]=(-2,\sqrt3)$

But:

$f^{-1}(0) = a \longrightarrow f(a)=0 \longrightarrow \begin{cases} 0 = 1-a \rightarrow a=1 \\ 0 = a^2 \longrightarrow a=0\\ \end{cases}$

$a=1$ is not a pre-image as $1>0$ then the interval $1-x$ of the function is not applied.

$a=0$ is not a pre-image as $f(0)=1$

From this I get that $f^{-1}(0)$ does not exist.

Then: $$f^{-1}([-2,3))=(-2,\sqrt3)-\{0\}$$

red = f(x), black = f^{-1}(x)

In the image it is shown the inverse of the function in the interval $-2\leq f(x)\leq 3$ which is in fact what I got, but the 0 is what I'm not sure about.

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Yes you way is right but we need also to include the $x=0$ value in the preimage indeed

$$f(0)=1 \in [-2,3)$$

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  • $\begingroup$ So $f^{-1}(0)$ exists? $\endgroup$ – user605734 MBS Nov 10 '18 at 13:35
  • $\begingroup$ @user605734MBS Yes I agree $f^{-1}(\{0\})=\emptyset$ (note that we need to use set notation inside) but $f(0)=1$ therefore $0$ is in the preimage and thus $$f^{-1}([-2,3))=(-2,\sqrt3)$$ $\endgroup$ – gimusi Nov 10 '18 at 13:40
  • $\begingroup$ This is because the function is continuous, right? $\endgroup$ – user605734 MBS Nov 10 '18 at 13:43
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    $\begingroup$ @user605734MBS No the finction is not continuous at $x=0$ but for $x=0$ the value for the function $f(0)=1$ is within the interval $[-2,3)$ and therefore also \${0\}$ is in the preimage of that interval. Revise careully the definition. $\endgroup$ – gimusi Nov 10 '18 at 13:46

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