- anonymous

Let , x^2+2x, x is > or = to -1.
Find its inverse and state its domain and range using interval notation. If required, use fractions, NOT decimals.
I did quadratic formula and got:
My answer: (-2+-sqrt(4+4x))/2
Domain of : (-oo,-1]
but this answer is wrong, what did I do wrong?

- chestercat

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- anonymous

I put [-1,oo) for range and that was correct...

- anonymous

Yeah, the range is \([-1,\infty)\).

- anonymous

\[\left( -2\pm \sqrt(2^2-4(1)(-1)\right)/2\]
This is my answer if thats clearer

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## More answers

- anonymous

hello anwar!
mistake is the \[\pm\] this a function so only one output for each input.

- anonymous

To find the inverse, set \(x=y^2+2y\), and try to solve for y. You will have \(y^2+2y-x=0\), by using the quadratic formula, we have:
\[y={-2\pm \sqrt{4+4x} \over 2}=-1\pm \sqrt{1+x}\]. But since the range is \((-1,\infty)\); we will take only \(+\sqrt{1+x}\).

- anonymous

Hello satellite!

- anonymous

hello. nice latex.

- anonymous

So, the inverse is \(\sqrt{1+x}-1\). Try to find the domain. And thank ssatellite :)

- anonymous

ok. S owhy would the answer for the domain not be (-oo,-1]?

- anonymous

The domain of the inverse has to be the same as the range of the original function. So, what do you think is the range of the original function, in the given domain \((-1,\infty)\)?

- anonymous

\([-1,\infty)\)*

- anonymous

why is \[\pm\] a mistake?

- anonymous

A function can't have a ±, because it would fail the vertical line test then.

- anonymous

The domain of the inverse, by the way, is \([-1,\infty)\).

- anonymous

oh, but I thought I'm to use quadratic formula to find the inverse of f(x)

- anonymous

in this situation

- anonymous

let me butt in for a second. think about \[f(x)=x^2\]
if you write \[x=y^2\] and solve for y you get
\[y=\pm\sqrt{x}\] but this is not a function.
and of course \[f(x)=x^2\] is not a one to one function. but if you restrict the domain of \[f(x)\] to x>0, then the inverse has to have range y > 0 and so you just write
\[f^{-1}(x)=\sqrt{x}\]
a perfectly good function.

- anonymous

I understand that there is only one output for one input, so how am i supposed to do that?

- anonymous

But, in the original function the domain is \([-1,\infty)\), and therefore the range of the inverse has to be \([-1,\infty)\). But if we take \(-\sqrt{x+1}\), that would lead to values that are not part of the range.

- anonymous

how do I express that?

- anonymous

It's already expressed :D

- anonymous

so, my question i guess is what is the range? [-1,oo)?

- anonymous

Yep.

- anonymous

Im confused. If it leads to wrong values, then why does it work?

- anonymous

Im confused. If it leads to wrong values, then why does it work?

- anonymous

That would be only If we take the minus the square as part of the solution, but we didn't. So, we're in the safe side.

- anonymous

OOOOO! I gotcha, thank you :)

- anonymous

can I ask you different question, about absolute value?
x=|y+3|

- anonymous

trying to figure out inverse of that?

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