anonymous
  • 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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I put [-1,oo) for range and that was correct...
anonymous
  • anonymous
Yeah, the range is \([-1,\infty)\).
anonymous
  • anonymous
\[\left( -2\pm \sqrt(2^2-4(1)(-1)\right)/2\] This is my answer if thats clearer

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anonymous
  • anonymous
hello anwar! mistake is the \[\pm\] this a function so only one output for each input.
anonymous
  • 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
  • anonymous
Hello satellite!
anonymous
  • anonymous
hello. nice latex.
anonymous
  • anonymous
So, the inverse is \(\sqrt{1+x}-1\). Try to find the domain. And thank ssatellite :)
anonymous
  • anonymous
ok. S owhy would the answer for the domain not be (-oo,-1]?
anonymous
  • 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
  • anonymous
\([-1,\infty)\)*
anonymous
  • anonymous
why is \[\pm\] a mistake?
anonymous
  • anonymous
A function can't have a ±, because it would fail the vertical line test then.
anonymous
  • anonymous
The domain of the inverse, by the way, is \([-1,\infty)\).
anonymous
  • anonymous
oh, but I thought I'm to use quadratic formula to find the inverse of f(x)
anonymous
  • anonymous
in this situation
anonymous
  • 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
  • anonymous
I understand that there is only one output for one input, so how am i supposed to do that?
anonymous
  • 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
  • anonymous
how do I express that?
anonymous
  • anonymous
It's already expressed :D
anonymous
  • anonymous
so, my question i guess is what is the range? [-1,oo)?
anonymous
  • anonymous
Yep.
anonymous
  • anonymous
Im confused. If it leads to wrong values, then why does it work?
anonymous
  • anonymous
Im confused. If it leads to wrong values, then why does it work?
anonymous
  • 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
  • anonymous
OOOOO! I gotcha, thank you :)
anonymous
  • anonymous
can I ask you different question, about absolute value? x=|y+3|
anonymous
  • anonymous
trying to figure out inverse of that?

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