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. > > > +---------------------------------------+ > / - - ---> 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 -----------------> > >