How to find the smallest root

Question

I have a continuous, differentiable, monotonic, bounded function called F[t]. If t -> Infinity then F[t] -> c. I know c, and I want to find the smallest t value called t*, that suffices F[t]=c (I am sure there is one).

Could somebody give me some hints about which functions to use for this task? Since I am afraid that FindRoot does not work in this case, because even if I know that there is a given T that T is smaller than t*, starting FindRoot from T as center won't give me the nearest solution, only a solution somewhere near T (Or am I wrong?). And neither does Solve work, because F[t] is not injective.

Answer 1

Briefly, one can use :

Reduce[ D[ F[x], x] == 0, x, Reals]

or if the function is not strictly monotonic one should select the maximal element out of the result in the boolean form, e.g.

Max @ (( List @@ Reduce[ D[ F[x], x] == 0, x, Reals] // Quiet)[[All, 2]])

The major problem is to provide some examples to demonstrate how it works. Since there have been none I will give the following :

Ex. 1. a class of $C^{1}$ functions

We define a family of differentiable i.e. $C^{1}$ functions F numbered with an eps parameter :

P[x_] := a x^3 + b x^2 + c x + d
W[x_, eps_] := P[x] //. Flatten @ Solve[{ #^2 == P[#], 1 == P[1], 2 # == P'[#], 1 == P'[1]}, 
                                        {a, b, c, d}]& @ (1 - eps)
Z[x_, eps_] := P[x] //. Flatten @ Solve[{ # == P[#], 2 == P[2], 1 == P'[#], 0 == P'[2]},
                                        {a, b, c, d}]& @ (2 - eps)
F[x_, eps_] := Piecewise[{ { x^2, 0 < x < 1 - eps}, 
                           { W[x, eps], 1 - eps <= x < 1}, 
                           { x, 1 <= x < 2 - eps}, 
                           { Z[x, eps], 2 - eps <= x < 2},
                           { 2, x >= 2} } ]

This definition is slightly involved because it is constructed with a few pieces of differentiable functions. The result fulfills the requirements. We introduced auxiliary functions P, W and Z to define F (monotonic, differentiable, and constant after x exceeds a certain point x0, (here x0 == 2)).
Now we can test the result for any eps (every eps > 0 define a differentiable function F[x, eps] ), e.g.

 Manipulate[ Plot[ F[x, eps], {x, 0, 2.3}, PlotRange -> {0, 2.3}] // Quiet, {eps, 0, 1}]

or showing a graph of F[x, eps] with a running parameter eps on the background of graphs for various eps :

    Animate[ 
        Show[ Plot[ Evaluate @ Table[ F[x, ep], {ep, 0, 1, 0.1}], {x, 0, 2.3}, 
                    PlotRange -> {0, 2.1}, ImageSize -> {650, 450}] // Quiet, 
              Plot[ F[x, eps], {x, 0, 2.3}, PlotRange -> {0, 2.1}, 
                    PlotStyle -> Thick, ImageSize -> {650, 450}] // Quiet, 
              Graphics[{ PointSize[0.015], Magenta,
                         Point @ {{1 - eps, F[1 - eps, eps]}, {2 - eps, F[2 - eps, eps]}}}]
            ],  {eps, 0, 1}]

Reduce[ D[F[x, 0.4], x] == 0, x, Reals] // Quiet
Max @ ( List @@ Reduce[ D[ F[x, 0.4], x] == 0, x, Reals] // Quiet)[[All, 2]]

 x <= 0 || x > 2.
 2.

The answer is x == 2.

Edit

We would like to test the method when the interesting point changes its location, therefore we add another class of functions to demonstrate the reliability of the approach based on Reduce.

Ex. 2. a class of $C^{\infty}$ functions

We construct a family of G functions numbered with an eps parameter, but this family is smooth (i.e. $C^{\infty}$) and the gluing point x == x0 is movable :

G[x_, eps_] := 
    Piecewise[{ {1 - eps^2/10 - eps Exp[-(1/(1 - 1/eps - x))], x < 1 - 1/eps},
                {1 - eps^2/10, x >= 1 - 1/eps} }]

This family of functions G is not analytic of course, (precisely, the only point where the functions are not analytic is x == 1 - 1/eps), however they are still smooth, i.e. we can find their derivatives of any order in x == 1 - 1/eps, for any eps, e.g.

Limit[ Table[ D[G[x, eps], {x, n}], {n, 1, 10}], x -> 1 - 1/eps, Direction ->  1]
Limit[ Table[ D[G[x, eps], {x, n}], {n, 1, 10}], x -> 1 - 1/eps, Direction -> -1]

{0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
{0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

thus its expansion in a Taylor series around x0 == 1 - 1/eps is constant unlike the function itself for x <= x0.

We can test that for any eps we get the correct result :

Manipulate[{Reduce[ D[ G[x, eps], x] == 0, x] // Quiet, 1 - 1/eps}, {eps, 0.1, 1}]

and we make a plot of a few functions :

GraphicsColumn[
    {Plot[ Evaluate @ Table[ G[x, eps], {eps, 0.2, 1.1, 0.1}], {x, -5, 2}, 
           PlotStyle -> Thick, PlotRange -> All, PerformanceGoal -> "Speed"], 

     Plot[ Evaluate @ Table[ G[x, eps], {eps, 0.5, 0.9, 0.1}], {x, -5, 2},  
           PlotStyle -> Thick, PlotRange -> {{-1.5, 0.1}, {0.9, 0.98}}, 
           PlotPoints -> 200, MaxRecursion -> 8]}]