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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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