The Inverse Function of f(x)=x^2+x

[Badballer]

I'm trying to work out the method of getting the inverse function of

$f(x) = x^2 + x$I already know the inverse but I would like to know the method used to obtain it. So far I have:

Made f(x) = y:
$y = x^2+2$

And then made y = x and x = y:
$x=y^2+y$

And then I did this but I'm not sure if it's correct:
$x=y(y+1)$
Apparently the solution is this
$x(y) = 1/2 (-1±sqrt(4 y+1))$

But I need to know the steps to get that. Hope you can help. :)

 

[micromass]

I'm trying to work out the method of getting the inverse function of

$f(x) = x^2 + x$


I already know the inverse but I would like to know the method used to obtain it. So far I have:

Made f(x) = y:
$y = x^2+2$

And then made y = x and x = y:
$x=y^2+y$

OK, let's leave from here. What you did afterwards is correct, but it won't help.
You have

$$y^2+y-x=0$$

this is a quadratic equation in y, so it can be solved with the quadratic formula. What does that give you?

 

[Badballer]

Oooh right so

a = 1
b = 1
c = -x

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

I think

Also sorry, I really fail with these itex tags.

 

[HallsofIvy]

By the way, since a real valued function of a real variable is "single valued", that does not have a true "inverse". What you are saying is that you can divide it into two functions, on either side of the vertex of the parabola, one having inverse function $f^{-1}(x)= (1/2)(-1+ \sqrt{4x+ 1})$ and the other having inverse $g^{-1}(x)= (1/2)(-1- \sqrt{4x+ 1})$

 

[zenose]

try this:

http://www.sosmath.com/algebra/quadraticeq/quadraformula/quadraformula.html

y(x) = a(x^2)+bx+c

0 = a(x^2)+bx+(c-y)

...

x(y) = (1/2a)(-b+(b^2 - 4a(c-y))^.5)

which is just the general quadratic with c replaced by (c-Y) so that x becomes a function of y. Abel showed this can't be done in general for polynomials with a finite number of arithmatic operations.

I would like to know how to the math symbols in something other than text. Is that what badballer was referring to in "itex tags"?

 

[Ayleeta]

I do not know the formula, but know the answer. Your function x^2+x looks like not-complete standart quadratic equation
a^2+2ab+b^2
(where a^2=x^2, 2ab=x, but missing b^2),
if it would be complete we could write it as
(a+b)^2.
To make it complete we add missing b^2,
if our a=x and 2ab=x, then our b is 1/2.
Our b^2 is 1/4, so we add it on both sides:
y+1/4=x^2+x+1/4.
y+1/4=(x+1/2)^2.
Square root both sides:
x+1/2=±sqrt(y+1/4)
move 1/2 from left side to right
x=±sqrt(y+1/4)-1/2.
Now you can transform right side of your equation to what you have
x=1/2(±2sqrt(y+1/4)-1).
Put 2 under square root:
x=1/2(±sqrt4(y+1/4)-1)
Then just do simple math

 

[SammyS]

I do not know the formula, but know the answer. Your function x^2+x looks like not-complete standart quadratic equation
a^2+2ab+b^2
(where a^2=x^2, 2ab=x, but missing b^2),
if it would be complete we could write it as
(a+b)^2.
To make it complete we add missing b^2,
if our a=x and 2ab=x, then our b is 1/2.
Our b^2 is 1/4, so we add it on both sides:
y+1/4=x^2+x+1/4.
y+1/4=(x+1/2)^2.
Square root both sides:
x+1/2=±sqrt(y+1/4)
move 1/2 from left side to right
x=±sqrt(y+1/4)-1/2.
Now you can transform right side of your equation to what you have
x=1/2(±2sqrt(y+1/4)-1).
Put 2 under square root:
x=1/2(±sqrt4(y+1/4)-1)
Then just do simple math

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