0
$\begingroup$

The question is-

Make the function $f:R\rightarrow$$R$, $f(x)=x^2$ invertible by making suitable adjustments to the domain and range.

So, here is what my teacher said-

Since we have to make the function onto, we must make the range equal to co-domain.

Range of y=$x^2$ is $[0\rightarrow \infty]$.

Also, the function must be one-one. So, the domain of the function must be ahead of the minima of the function.

So, finally the function becomes $f:[0\rightarrow \infty]\rightarrow [0 \rightarrow \infty]$.

So, the graph looks like thisenter image description here

But, why isn't this part of the graph also included in the answer enter image description here

$\endgroup$

2 Answers 2

0
$\begingroup$

Your goal was to restrict $f$ to a subset of $\mathbb{R}$ in which it becomes a bijection i.e injective and surjective. If you include that part of the graph then $[- \epsilon, \epsilon] \in \textbf{dom}(f)$ for some $\epsilon >0$. Hence, you have $f(- \epsilon) = f(\epsilon)$ and so $f$ is not injective.

$\endgroup$
2
  • $\begingroup$ But, I am including only $[-\infty \rightarrow 0]$. $\endgroup$ Commented May 14, 2017 at 13:20
  • 1
    $\begingroup$ Oh, I wasn't clear on that. Yes, $f|_{(- \infty,0]}$ is also invertible. $\endgroup$ Commented May 14, 2017 at 13:38
0
$\begingroup$

There are many more ways to make that function invertible.

Take any subset $X$ of the real line such that $x$ and $-x$ aren't included simultaneously, and let $F$ be the image of this subset. Then $f:X\to Y$ is invertible.

This includes $$f:(-\infty,0]\to[0,\infty),$$ but also more exotic definitions such as $$f:[-2,-1]\cup(3,5)\to[1,4]\cup(9,25)$$

or

$$\bigcup_{k=0}^\infty[2k,2k+1)\cup(-2k-2,-2k-1]\to[0,\infty).$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .