# inverse function not working -- but providing values for some arguments, a symbol otherwise?

Could you help me with what Mathematica is doing here, in the last lines? (The rest is here for context.) I removed the output to let you simply run this. Thanks.

k=10/3
u[c_,l_]:=Log[c]-Log[1+l^(1+k)/(1+k)]
T[z_]:=(1-0.84/1.3) * z
lType[n_]:=ArgMax[{u[n l-T[n l],l],l>=0},l]
zType[n_] := n lType[n]
Type := InverseFunction[zType]
Type
Type
Type


With a much longer version with some context, you can also switch to https://stackoverflow.com/questions/10641713/mathematica-define-inverse-only-where-it-exists

Or please reconsider the same code with some output to see what my problem is with the inverse:

k=10/3
10/3
Utility function as in Saez 2001 allowing for income effects:
u[c_,l_]:=Log[c]-Log[1+l^(1+k)/(1+k)]
Actual tax schedule (approximate)
T[z_]:=(1-0.84/1.3) * z
Equation 14:
lType[n_]:=ArgMax[{u[n l-T[n l],l],l>=0},l]
zType[n_] := n lType[n]
Type := InverseFunction[zType] does not seem to work.
Type := InverseFunction[zType]
Type
5030.985810589009
Type
(zType^(-1))
Type
(zType^(-1))

• It might help if you could re-post your code so that it is easily copied and pasted into Mathematica (i.e. without the Ins] and with the Outs separated. A bit more context might help too. May 17, 2012 at 19:16
• @ian.milligan: Thanks, I'll try. This is how you can copy from Mathematica, no? Copy as plain text? On context I could guide you to a similar StackOverflow posting, which did not go anywhere partly because of the context. :) And I was embarrassed by the cross-posting, a no-no. May 17, 2012 at 19:27
• László, if you do a Merge Cells before the copy you won't have all the In/Out tags. May 17, 2012 at 19:42
• @Mr.Wizard: This is great, what a newbie I am! Thanks. I hope all would run just as well this way? May 17, 2012 at 19:56
• @László Even better: use this to copy code meta.mathematica.stackexchange.com/a/155/745 May 21, 2012 at 7:49

 lType[n_]:=ArgMax[{u[n l-T[n l],l],l>=0},l]

l` is not defined.