I want to find the root of a function that has a piecewise part. This means that the value Mathematica will search will determine in which piece of the function we will be. I don't my example is the best, but this is the best example I could think of (and thank you @m_goldberg for helping be to Inactivate the function):

f[a_?NumericQ] := FindRoot[2 x^2 + Log[x] - a, {x, 0.01}][[1, 2]]
f[a_] := Inactivate[FindRoot[2*x^2 + Log[x] - a, {x, 0.01}][[1, 2]], 
  FindRoot | Part]
g[a_] := Piecewise[{{1 - f[a] + Log[f[a]], 
    a <= 1}, {Log[2*f[a]] + f[a], a > 1}}]
Findroot[g[a] + Log[a], {a, 0.1, 0.01, 10}]

Any ideas?


1 Answer 1


I would define g using NumericQ just as you did for f instead of using Inactivate:

f[a_?NumericQ] := FindRoot[2 x^2+Log[x]-a, {x,0.01}][[1,2]]
g[a_?NumericQ] := Piecewise[{

Then try using FindRoot:

FindRoot[g[a] + Log[a], {a,0.1,0.01,10}]

FindRoot::lstol: The line search decreased the step size to within tolerance specified by AccuracyGoal and PrecisionGoal but was unable to find a sufficient decrease in the merit function. You may need more than MachinePrecision digits of working precision to meet these tolerances.

{a -> 1.}

Let's plot your expression over the region of interest:

Plot[g[a] + Log[a], {a, 0, 10}]

enter image description here

The FindRoot message occurs because the root occurs at the piecewise boundary.

  • $\begingroup$ Thank you very much, @Carl Woll. Indeed my example was not the best because of this boundary issue, but on my problem this is not the case. Thank you again! $\endgroup$
    – Laura K
    Oct 29, 2017 at 18:30

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