# Thread: Why isn't the other part of the graph taken?

1. ## Why isn't the other part of the graph taken?

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 this

[1]: https://i.stack.imgur.com/OkkHg.jpg

But, why isn't this part of the graph also included in the answer

[2]: https://i.stack.imgur.com/Y0S6R.jpg

2. ## Re: Why isn't the other part of the graph taken?

It cannot be "included" in the answer because the two together give all of $\displaystyle y= x^2$ which is not invertible. Both -2 and 2 are mapped to 4 so there is not unique inverse for 4. However, that is an alternate answer. "$\displaystyle y= x^2$ for $\displaystyle x\ge 0$" has inverse $\displaystyle y= \sqrt{x}$ while "$\displaystyle y= x^2$ for $\displaystyle x\le 0$" has inverse $\displaystyle y= -\sqrt{x}$. If fact, there are infinitely many correct answers to this problem. "$\displaystyle y= x^2$ for $\displaystyle x> 1$" is invertible and so is "$\displaystyle y= x^2$ for $\displaystyle x< -2$".