       • 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.
>
>
>    +---------------------------------------+
>    /   - - ---> C y b e r M a n <--- - -   /
>    /    _______ Milano -- ITALY _______    /
>    /->>> e-mail:  cyberman at zerocity.it <<<-/
>    /          PGP KeyID: 487C6475          /
>    +---------------------------------------+
>         - ----< Key fingerprint >---- -
> 8F 50 99 35 BF A4 83 6C 68 11 C9 31 DC A7 65 F2
>       <---------------- at ----------------->
>
>

```

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