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anonymous
 5 years ago
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?
anonymous
 5 years ago
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?

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I put [1,oo) for range and that was correct...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, the range is \([1,\infty)\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\left( 2\pm \sqrt(2^24(1)(1)\right)/2\] This is my answer if thats clearer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hello anwar! mistake is the \[\pm\] this a function so only one output for each input.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0To find the inverse, set \(x=y^2+2y\), and try to solve for y. You will have \(y^2+2yx=0\), 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
 5 years ago
Best ResponseYou've already chosen the best response.0So, the inverse is \(\sqrt{1+x}1\). Try to find the domain. And thank ssatellite :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok. S owhy would the answer for the domain not be (oo,1]?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0why is \[\pm\] a mistake?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A function can't have a ±, because it would fail the vertical line test then.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The domain of the inverse, by the way, is \([1,\infty)\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, but I thought I'm to use quadratic formula to find the inverse of f(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let 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
 5 years ago
Best ResponseYou've already chosen the best response.0I understand that there is only one output for one input, so how am i supposed to do that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But, 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
 5 years ago
Best ResponseYou've already chosen the best response.0how do I express that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's already expressed :D

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, my question i guess is what is the range? [1,oo)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Im confused. If it leads to wrong values, then why does it work?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Im confused. If it leads to wrong values, then why does it work?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That 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
 5 years ago
Best ResponseYou've already chosen the best response.0OOOOO! I gotcha, thank you :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can I ask you different question, about absolute value? x=y+3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0trying to figure out inverse of that?
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