$\displaystyle f : R-> R$

$\displaystyle f(x)= x^{2}+2x+3$

a) Define a restriction of $\displaystyle f$ that admits inverse.

Can u help me ?? I dont have any idea, how to do that.

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- Nov 26th 2011, 03:13 PMFabio010Function restriction.
$\displaystyle f : R-> R$

$\displaystyle f(x)= x^{2}+2x+3$

a) Define a restriction of $\displaystyle f$ that admits inverse.

Can u help me ?? I dont have any idea, how to do that. - Nov 26th 2011, 03:35 PMskeeterRe: Function restriction help plz
- Nov 26th 2011, 03:36 PMAlso sprach ZarathustraRe: Function restriction help plz
- Nov 26th 2011, 03:51 PMFabio010Re: Function restriction help plz
I already know the inverse function.

$\displaystyle x = +/-\sqrt{y-2}-1$

The inverse graph is a parabole with vertex(2,-1)

I dont understood the part f(x) is 1-1

btw: Sorry for posting in wrong section(this problem is from a university calculus test) - Nov 26th 2011, 04:07 PMemakarovRe: Function restriction help plz
A function is 1-1 (or injective) if it maps different arguments to different values. For a function to have an inverse, it must be injective: if $\displaystyle y = f(x_1) = f(x_2)$ where $\displaystyle x_1\ne x_2$, then $\displaystyle f^{-1}(y)$ is not defined because it cannot be both $\displaystyle x_1$ and $\displaystyle x_2$. Therefore, you need to find a restriction where the function is 1-1, as skeeter said.

- Nov 26th 2011, 04:16 PMFabio010Re: Function restriction help plz
Hum....so to be injective:

we know that vertex of f is (-1,2) so,

the restriction is [-1, +oo[ ??? as skeeter said... - Nov 26th 2011, 04:22 PMemakarovRe: Function restriction help plz
- Nov 26th 2011, 04:36 PMFabio010Re: Function restriction help plz
Ok, so to f admit inverse we must restrict the domain in order to make f injective. (Sorry my english)

- Nov 26th 2011, 04:37 PMemakarovRe: Function restriction help plz
Yes, that's right.