Function restriction.

• Nov 26th 2011, 04:13 PM
Fabio010
Function restriction.
$f : R-> R$

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

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

Can u help me ?? I dont have any idea, how to do that.
• Nov 26th 2011, 04:35 PM
skeeter
Re: Function restriction help plz
Quote:

Originally Posted by Fabio010
$f : R-> R$

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

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

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

$f(x) = x^2 + 2x + 3 = x^2 + 2x + 1 + 2 = (x+1)^2 + 2$

note that the graph of f(x) is a parabola with vertex at (-1,2)

for f(x) to have an inverse, the domain must be restricted so that f(x) is 1-1 ... that would be values of x such that x > -1
• Nov 26th 2011, 04:36 PM
Also sprach Zarathustra
Re: Function restriction help plz
Quote:

Originally Posted by Fabio010
$f : R-> R$

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

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

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

f(x)=y=x^2+2x+3

Start by solving x^2+2x+(3-y)=0
• Nov 26th 2011, 04:51 PM
Fabio010
Re: Function restriction help plz
I already know the inverse function.

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

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, 05:07 PM
emakarov
Re: 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 $y = f(x_1) = f(x_2)$ where $x_1\ne x_2$, then $f^{-1}(y)$ is not defined because it cannot be both $x_1$ and $x_2$. Therefore, you need to find a restriction where the function is 1-1, as skeeter said.
• Nov 26th 2011, 05:16 PM
Fabio010
Re: 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, 05:22 PM
emakarov
Re: Function restriction help plz
Quote:

Originally Posted by Fabio010
so to be injective:

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

the restriction is [-1, +oo[ ???

Yes, or ]-oo, -1], or any subset of those two.
• Nov 26th 2011, 05:36 PM
Fabio010
Re: 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, 05:37 PM
emakarov
Re: Function restriction help plz
Yes, that's right.