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Re: Help me please!!!
*To*: mathgroup at smc.vnet.net
*Subject*: [mg5814] Re: [mg5770] Help me please!!!
*From*: Robert Pratt <rpratt at math.unc.edu>
*Date*: Wed, 22 Jan 1997 00:44:17 -0500
*Sender*: owner-wri-mathgroup at wolfram.com
First, you can graph the parabola and convince yourself that the function
is one-to-one (hence invertible) for x>=0.
Now to find the formula for the inverse, switch x and y and solve for y
by completing the square:
y=x^2+x (original function)
x=y^2+y (switching x and y)
x+(1/2)^2=y^2+y+(1/2)^2 (adding the "magic" quantity (1/2)^2 to both
sides to complete the square)
x+1/4=(y+1/2)^2 (simplifying)
Sqrt[x+1/4]=y+1/2 (taking the positive square root)
y=Sqrt[x+1/4]-1/2 (solving for y)
Note that y>=0 if x>=0.
This procedure gives the correct formula for the inverse, as can be
checked by computing the composition of the first with the second and the
second with the first:
Let f(x)=x^2+x, x>=0
and g(x)=Sqrt[x+1/4]-1/2
Check that f[g(x)]=x and g[f(x)]=x.
Rob Pratt
Department of Mathematics
The University of North Carolina at Chapel Hill
CB# 3250, 331 Phillips Hall
Chapel Hill, NC 27599-3250
rpratt at math.unc.edu
On Sat, 18 Jan 1997, Cyberman wrote:
> Hello. I've a little problem:
>
> I must invert this function: y = (x^2) + x for x >= 0
>
> Is there anybody that could help me?
>
> Thanx.
>
>
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