Thread: Function restriction.

1. Function 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.

2. Re: Function restriction help plz

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

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

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

Can u help me ?? I dont have any idea, how to do that.
$\displaystyle f(x) = x^2 + 2x + 3 = x^2 + 2x + 1 + 2 = (x+1)^2 + 2$

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

3. Re: Function restriction help plz

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

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

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

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

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

Start by solving x^2+2x+(3-y)=0

4. Re: 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)

5. 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 $\displaystyle y = f(x_1) = f(x_2)$ 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.

6. 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...

7. Re: Function restriction help plz

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.

8. 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)

9. Re: Function restriction help plz

Yes, that's right.