Maths help! (Inverse)
Hey guys, I was looking for a little bit of help.
How do you complete this equation?
Consider the function g: [-1,infinite) -> R, g(x) = x^2 +2x
a. Find g^-1 , stating the domain and range.
b. Sketch the graph of g^-1
Thanks in advance! =)
January 27, 2011
13 Comments • Newest first
Math!
Hissssssss
[quote=PhaseOfFact]@above
x can equal -1, and y can also equal -1. You need "greater than or equal to", but I dunno how to type that symbol lol[/quote]
wouldn't it be <= or >= ? as in C++? can't think of this symbol to be written differently
OHHH!
Haha it makes so much sense now, thanks everybody!
I did it like this (mostly the same as hyperfire)
1. y = x^2 + 2x
2. x = y^2 + 2y
3. x + 1 = y^2 +2y + 1
4. x + 1 = (y+1)^2
5. sqrt(x + 1) = y + 1
6. sqrt(x+1) - 1 = y
@above
I know, just making sure kunannnn didn't forget the "greater than or equal to" part xD
[quote=PhaseOfFact]@above
x can equal -1, and y can also equal -1. You need "greater than or equal to", but I dunno how to type that symbol lol[/quote]
Which is exactly why I didn't mention it. You can't type that symbol.
[quote=hyperfire7]To find the inverse, switch x and y of the original function. Then solve for x to get the inverse equation.
1) Y = X^2 + 2X
2) X = Y^2 + 2Y
3) Y(Y+2) = -X
4) Y = (Sqrt (X+1)) - 1
With that in mind, Domain : -1 < X < Infinite, since a negative square root is a non-real answer.
Range would be -1 < Y < Infinite.[/quote]
How did you get from 3 to 4? o.o
@TheReseacher
It's the beginning of year 12. I haven't started school yet =/, I'm sure my teacher could explain it but I wanted
to get a headstart.
@above
x can equal -1, and y can also equal -1. You need "greater than or equal to", but I dunno how to type that symbol lol
To find the inverse, switch x and y of the original function. Then solve for x to get the inverse equation.
1) Y = X^2 + 2X
2) X = Y^2 + 2Y
3) Y(Y+2) = -X
4) Y = (Sqrt (X+1)) - 1
With that in mind, Domain : -1 < X < Infinite, since a negative square root is a non-real answer.
Range would be -1 < Y < Infinite.
What grade is this work?
Yea, so for b, just graph that equation xD
Shouldn't be too hard~
Wouldn't the inverse just mean switching the x and y, and then solving for x? Thats what I did...
well, for a. i think u must combine the function g, like gog, OR g(g(x))... that's how u'll find g^-1..and the domain and range...well...there would be a second way to find out g^-1, and using that way the domain and range swap places...but here, i actually can't think :S
Im pretty sure it would be
a. g^-1(x) = sqrt(x+1)-1 Domain: [-1,infinite) Range: [-1, infinite)
b. Can't really graph on basil lol