If x belongs to (0, π / 2), find the minimum value of 2tanx + 1 / TaNx

If x belongs to (0, π / 2), find the minimum value of 2tanx + 1 / TaNx

According to X ∈ (0, π / 2)
So TaNx ∈ (0, + ∞)
Suppose t = TaNx ∈ (0, + ∞)
therefore
2t+1/t>=2√(2t*1/t)=2√2
So, the minimum value is 2 √ 2
When 2T = 1 / T
2t^2=1
When t = √ 2 / 2, take the equal sign, that is, x = π / 4

If x = (0,4], then the maximum value of 2tanx - (1 / TaNx) is?

Because (0, π) is contained in (0,4]
So TaNx ∈ (- ∞, + ∞)
Let y = TaNx
2y-(1/y)
When y - > + ∞, 2Y - (1 / y) - > + ∞
So the maximum does not exist

Given the function f (x) = {- √ 3 (TaNx) & # 178; + 2tanx + √ 3} / {(TaNx) & # 178; + 1}, X belongs to R
(1) Finding the minimum positive period of function f (x) and a symmetry axis
(2) If x belongs to [π / 4,3 π / 4], find the minimum and maximum of F (x)
(3) If G (x) = f (x) - A and X belongs to (- π / 12,2 π / 3], when there are 0, 1 and 2 solutions for G (x) = 0, the value range of corresponding a is obtained respectively?

Given the function f (x) = [- (√ 3) Tan & # 178; X + 2tanx + # 3] / (Tan & # 178; X + 1), X belongs to R
(1) Finding the minimum positive period of function f (x) and a symmetry axis
(2) If x belongs to [π / 4,3 π / 4], find the minimum and maximum of F (x)
(3) If G (x) = f (x) - A and X belongs to (- π / 12,2 π / 3], when there are 0, 1 and 2 solutions for G (x) = 0, the solutions for G (x) = 0 are obtained
The value range of a
(1).f(x)=[-(√3)tan²x+2tanx+√3]/(tan²x+1)=[-(√3)tan²x+2tanx+√3]/(sec²x)
=-(√3)sin²x+2sinxcosx+(√3)cos²x=sin2x+(√3)(cos²x-sin²x)=sin2x+(√3)cos2x
=2[(1/2)sin2x+(√3/2)cos2x]=2[sin2xcos(π/3)+cos2xsin(π/3)]=2sin(2x+π/3)
So its minimum positive period T = π; let 2x + π / 3 = π / 2, then 2x = π / 2 - π / 3 = π / 6, so x = π / 12 is its axis of symmetry
(2) If x ∈ [π / 4,3 π / 4], then f (x) has the minimum value minf (x) = f (7 π / 12) = 2Sin (7 π / 6 + π / 3) when x = 7 π / 12
=When x = π / 4, the maximum value of F (x) is MAXF (x) = f (π / 4) = 2Sin (π / 2 + π / 3) = 2Sin (5 π / 6)
=2sin(π-π/6)=2sin(π/6)=2×(1/2)=1.
(3).g(x)=2sin(2x+π/3)-a=0.①,(-π/12