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

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