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I found an example of FindRoot usage that I couldn't find documented. It looks like:

FindRoot[f[x] == 0.4, {x, 0, 1, 0, 1}]

The documentation shows this case FindRoot[lhs==rhs, {x, xstart, xmin, xmax}] searches for a solution, stopping the search if x ever gets outside the range. But the example I found has an extra argument. What does it mean?


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I put True in that position and got the error FindRoot::bbound: Search region bound True for variable number 1 is not a number or Infinity. >> so my guess is search region bound. FindRoot[Cos[x] == 0.4, {x, 0, 1, 0, 3}] finds the solution $x \approx 1.59$. Another odd thing FindRoot[Cos[x] == 0.4, {x, 0, 1, 0, 0.5}] FindRoot::streg: The starting point {1.} is not in the search region {{0.},{0.5}}. >> – ssch Jan 7 '13 at 15:41
I think the 4 values are : 2 starting points and the search region; 2 starting points are used by the "Secant" method for instance. For instance try FindRoot[x^2 - 2 == 0, {x, 1.1, 1.9, 1.2, 2}, Method -> "Secant"] and move around the 1.1 or the 1.9. – b.gatessucks Jan 7 '13 at 15:43
@b.gatessucks that agrees with my last error. – ssch Jan 7 '13 at 15:44

As suggested by b.gatessucks, the 4 values are the two starting points used by the "Secant" method for instance and the search region. In fact if you use

FindRoot[f[x], {x, x_a, x_b, -Infinity, Infinity}]

you get always the same result as for

FindRoot[f[x], {x, x_a, x_b}],

whilst if you take

FindRoot[f[x], {x, x_a, x_b, -y, y}]

with x_a or x_b bigger than y in modulus you'll get the error

FindRoot::streg: "The starting point is not in the search region"
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