How does one check a function monotonicity?

Question

the question is to analyze $f(x)$ for monotonicity. I know I need to use D[f[x], x] and check if it is less than or greater to zero, but NSolve is not working properly.

f[x_] := 
  Sqrt[1 - a + (b + c) (c + b x + c x) - 
    Sqrt[(b + 2 c + a c)^2 - (b^2 - c^2)(3 + 4 a - a^2 + b^2 - c^2) x + 
      (b^2 - c^2) x^2]] + 
  Sqrt[1 - a + (b + c) (c + b x + c x) + 
    Sqrt[(b + 2 c + a c)^2 - (b^2 - c^2) (3 + 4 a - a^2 + b^2 - c^2) x + 
     (b^2 - c^2) x^2]] + 
  Sqrt[1 + a - (b - c) (c - b x - c x) - 
    Sqrt[(b - 2 c - a c)^2 - (b^2 - c^2) (3 - 4 a - a^2 + b^2 - c^2) x + 
      (b^2 - c^2) x^2]] + 
  Sqrt[1 + a - (b - c) (c - b x - c x) + 
    Sqrt[(b - 2 c - a c)^2 - (b^2 - c^2) (3 - 4 a - a^2 + b^2 - c^2) x + 
      (b^2 - c^2) x^2]]

Reduce[
  0 <= (1/4) (1 - a - b - c) <= 1 &&
  0 <= (1/4) (1 + a + b - c) <= 1 &&
  0 <= (1/4) (1 + a - b + c) <= 1 &&
  0 <= (1/4) (1 - a + b + c) <= 1 &&
  a^2 < b^2 < c^2, 
  {a, b, c}, Reals];

NSolve[D[f[x], x] > 0]

I did try to do this im many ways. I'm pretty new to Mathematica and today I thought of doing it this way, but it's not working

Answer 1

You may use Simplify.

If $f(x)$ is a function, then

Simplify[f[x]>=f[y], x>=y]

should evaluate to the necessary condition for the monotonicity to hold. If necessary, you can use FullSimplify

Answer 2

As an example, consider this function g and its derivative gd:

g[x_] := x - 1/10*x^3;
gd[x_] = D[g[x], x];

Now you want to study monotonicity in a certain interval. For this, you will need to find the local extrema of your function. The corresponding x-values of your extrema can be found using Solve, NSolve or FindRoot (for this simple example, Solve works - in general the other two approaches are required for numerical solutions).

xextreme = Solve[gd[x] == 0, x][[;; , 1, 2]]
{-Sqrt[(10/3)], Sqrt[10/3]}

Now you need to study the derivative's sign in the intervals separated by these values. To do so, we generate a list that has values "left" and "right" as well as in between the values found in xextreme. This is general, though very likely not the most beautiful approach (but the first that came to my mind and suffices).

testvals = Append[Prepend[(Drop[Riffle[#, # + 1/2*Append[Differences@#, 0]], -1] &@xextreme), -2*#], 2*#] &@Max[Abs[xextreme]]
{-2 Sqrt[10/3], -Sqrt[(10/3)], 0, Sqrt[10/3], 2 Sqrt[10/3]}

Now check the sign of the derivative:

TableForm[Transpose[{testvals, Sign[gd[#]] & /@ testvals}], TableHeadings -> {None, {"x", "sign(g'(x))"}}]

• Negative sign: decreasing function
• Positive sign: increasing function
• Zero: local extrema (including possibility of saddle point)

Note that if you use NSolve or FindRoot, you should rather use Sign[Chop@gd[#]] inside the TableForm line to get around numerical issues.

Update

To be very specific for your function in question: Have a look at this beautiful package and evaluate these lines

Needs["ConstantsGrouping`"];
ClearAll[fc,fcs, rule1,rule2];
{fc[x_], rule1} = GroupConstants[f[x], x, GeneratedParameters -> (Symbol["kf" <> ToString[#]] &)];
{fcs[x_], rule2} = GroupConstants[Numerator@Together@D[fc[x], x], x, GeneratedParameters -> (Symbol["kfs" <> ToString[#]] &)];

Then, observe the "beauty" of fcs[x]. All these hundreds of constants kxyz depend on a,b,c as specified in rule1,rule2. You may even do fcs[x]/.rule2/.rule1 and see how large your expression becomes. Note that this is only the numerator of the derivative f'[x]. You would need to find roots of this thing, which appears to be impossible symbolically. Also, the numerical values will all (AFAIK) fail unless you specify values for a,b,cbecause there will be non-numerical values inside your expressions.