Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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 \(+\sqrt{1+x}\).

Answer 6

Hello satellite!

Answer 7

hello. nice latex.

Answer 8

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

Answer 9

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

Answer 10

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 \((-1,\infty)\)?

Answer 11

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

Answer 12

why is \[\pm\] a mistake?

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

in this situation

Answer 17

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.

Answer 18

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

Answer 19

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.

Answer 20

how do I express that?

Answer 21

It's already expressed :D

Answer 22

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

Answer 23

Yep.

Answer 24

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

Answer 25

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

Answer 26

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.

Answer 27

OOOOO! I gotcha, thank you :)

Answer 28

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

Answer 29

trying to figure out inverse of that?