Pre-Image of a non continuous function.

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

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.

1 Answer

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

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

• So $f^{-1}(0)$ exists? – user605734 MBS Nov 10 '18 at 13:35
• @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 $$f^{-1}([-2,3))=(-2,\sqrt3)$$ – gimusi Nov 10 '18 at 13:40
• This is because the function is continuous, right? – user605734 MBS Nov 10 '18 at 13:43
• @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. – gimusi Nov 10 '18 at 13:46