Given that x ∈ [- π 3, π 4], f (x) = tan2x + 2tanx + 2, find the maximum and minimum of F (x), and find the corresponding x value
f(x)=tan2x+2tan x+2=(tan x+1)2+1. … 2∵x∈[-π3,π4],∴tan x∈[-3,1]. … When Tan & nbsp; X = - 1, that is, x = - π 4, y has a minimum value, Ymin = 1 When Tan & nbsp; X = 1, that is, x = π 4, y has a maximum value, ymax = 5 twelve