• We’re currently investigating an issue related to the forum theme and styling that is impacting page layout and visual formatting. The problem has been identified, and we are actively working on a resolution. There is no impact to user data or functionality, this is strictly a front-end display issue. We’ll post an update once the fix has been deployed. Thanks for your patience while we get this sorted.

help please! math question!

O

Omegachi

Diamond Member
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)
 
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
 
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?
 
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.

 
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.
Click to expand...


hmm, doing it this way, i ended up with

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

what do i do from there?
 
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?
 
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?
 
You must log in or register to reply here.

TRENDING THREADS

Back
Top