help please! math question!

[Omegachi]

how do you find the inverse function of this? I really can't figure it out myself please help

h(x)= x + sqrt(x)

find (h^(-1))(x)

 

[Schadenfreude]

Quick answer: find out yourself!

 

[Omegachi]

awww, i can't figure it out

hints?

 

[Omegachi]

bump... can anyone at least help me find

y= x + sqrt(x)

in terms of x = ?

 

[Omegachi]

bump?

 

[TuxDave]

Heh... if I find it in terms of x=? Doesn't that just give away the answer? Anyways, here's a hint, try completing the square on the right side. Here's another hint, 1/4 is an important number

 

[Whitecloak]

solve for x in terms of y. Hint: it will be a quadratic eqn if you square both sides

 

[Omegachi]

hmm... so i square both sides...

i get

y^2 = (x + sqrt(x)) * (x + sqrt(x))

>> y^2 = x^2 + 2x*sqrt(x) + x

>> y^2 = x( x +2*sqrt(x) + 1)

.... so what do i do now?

 

[Ophir]

I don't think so. It's more like: x= y + sqrt (y), then solve for y = ?

 

[Whitecloak]

take x to the LHS so that you get

y - x = sqrt(x)

Now square both sides & solve for x.

You will get the inverse function.

 

[Omegachi]

Originally posted by: Ophir
I don't think so. It's more like: x= y + sqrt (y), then solve for y = ?

dude, its the same thing.

 

[Omegachi]

Originally posted by: whitecloak
take x to the LHS so that you get

y - x = sqrt(x)

Now square both sides & solve for x.

You will get the inverse function.

hmm, doing it this way, i ended up with

y^2 - 2xy = x^2 - x

what do i do from there?

 

[Ophir]

Originally posted by: Omegachi

Originally posted by: Ophir
I don't think so. It's more like: x= y + sqrt (y), then solve for y = ?

dude, its the same thing.

NM. I don't remember how to do this. Sorry for any confusion

 

[Whitecloak]

solve it for x

 

[Omegachi]

wow wait, after i did that, i completed the square and crap like that. and i ended up with

x= ((1 + y)^2 + 1)/5

and that is not the answer.... close, but not quite

what did i do wrong?

 

[Whitecloak]

dude, you solved the equation incorrectly

y = x - sqrt(x)
y - x = sqrt(x)
y^2 - 2xy + x^2 = x

ie
x^2 - (2y+1)x + y^2 = 0

Therefore,

x = {(2y+1) (+/-) sqrt[(2y+1)^2 - 4y^2]}/2

x = {2y+1 (+/-) sqrt (4y+1)}/2

what do you mean by (h^(-1))(x) anyway?

 

[Omegachi]

ohhh, i figured it out thanks all!!!!