How can I find the domain of $x$ for which the function $g(x)geq0;$?

Question

I have this function ($x>0$)
$$f (x)=frac{sqrt{g (x)}+4 x left(x^2+1right) sin (pi  x) cos ((3+pi ) x)}{x^4+2 x^2+1+left(4 x^2+left(x^2-1right)^2 cos (2 pi  x)right)}$$
f[x_] := (
  Sqrt[g[x]] + 4 x (1 + x^2) Cos[(3 + π) x] Sin[π x])/(
  1 + 2 x^2 + x^4 + (4 x^2 + (-1 + x^2)^2 Cos[2 π x]));

where $g(x)=4 x^2+left(x^2-1right)^2 cos (2 pi  x)-left(x^2+1right)^2 cos (2 (3+pi ) x);$.
g[x_] := 4 x^2 + (-1 + x^2)^2 Cos[2 π x] - (1 + x^2)^2 Cos[
     2 (3 + π) x];

I want to check the range of function $f(x)$ for those values of $x$ in which $g(x)geq0;$.
-How can I find the domain of $x$ for which $g(x)geq0;$?

Ulrich Neumann · Answer

Try ( Thanks @SjoerdSmit helpful comment )

cond=FunctionDomain[{f[x], g[x] >= 0 }, x, Reals] //FullSimplify
(*1 + 6 x^2 + x^4 + (-1 + x^2)^2 Cos[2 [Pi] x] != 0 && 
4 x^2 + (-1 + x^2)^2 Cos[2 [Pi] x] >= (1 + x^2)^2 Cos[2 (3 + [Pi]) x]*)

This condition gives (Thanks @user64494 )

Reduce[N[1 + 6 x^2 + x^4 + (-1 + x^2)^2 Cos[2 [Pi] x] != 0 &&      4 x^2 +(-1 + x^2)^2 Cos[2 [Pi] x] >= (1 + x^2)^2 Cos[        2 (3 + [Pi]) x]] && x>= 0 && x <= 10, x, Reals]
(*0 <= x <= 0.386763 || 0.590121 <= x <= 1.02279 || 
1.02333 <= x <= 1.47324 || 1.60461 <= x <= 2.03514 || 
2.06742 <= x <= 2.43253 || 2.67214 <= x <= 3.049 || 
3.11778 <= x <= 3.42438 || 3.70494 <= x <= 4.06349 || ...<= x <= 10.*)

visualization:

Plot[1, {x, 0, 10}, 
RegionFunction ->Function[{x},1 + 6 x^2 + x^4 + (-1 + x^2)^2 Cos[2 [Pi] x] != 0 &&4 x^2 + (-1 + x^2)^2 Cos[2 [Pi] x] >= (1 + x^2)^2 Cos[2 (3 + [Pi]) x]]]

How can I find the domain of $x$ for which the function $g(x)geq0;$?

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