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FindROot and substitutions


Hi,

I have found a workaround for this problem, but I would like to
understand how Mathematica wants you to think so I'm asking the
question anyway.

Why does "bad" not work, while "good" works?

g := x z;
bad[z_] := FindRoot[ g == 5, {x, -10, 10 }];
bad[5]
good[k_] := FindRoot[ g == 5 /. z -> k, {x, -10, 10 }];
good[6]

Thanks!

Aaron


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