三角題目兩題 (求函數定義域及值域)

(1) The function is defined if and only if 1-tanx ≧ 0. So
tanx ≦ 1.
The general solution for this inequality is
nπ - π/2 ≦ x ≦ nπ +π/4 , where n is any integer.
Hence the domain is {[nπ - π/2, nπ +π/4] : n is any integer}
For x is this domian, 1-tanx ranges from 0 to infinity. Hence the range of y is [0,∞).

(2) √(tanx - tan2x) is defined when tanx - tan2x ≧ 0, i.e.,
0 ≦tanx ≦1
Hence the function is defined for nπ ≦ x < nπ +π/4, n is any integer.
The domain is {[nπ,nπ +π/4] : n is any integer}
For x in this domain, 0 ≦tanx ≦1, 0 ≦ tanx - tan2x ≦ 1/4, hence the range of y is
[0,1/2].

1) The function is defined if and only if 1-tanx ≧ 0. So
tanx ≦ 1.
The general solution for this inequality is
nπ - π/2 ≦ x ≦ nπ +π/4 , where n is any integer.
Hence the domain is {[nπ - π/2, nπ +π/4] : n is any integer}
For x is this domian, 1-tanx ranges from 0 to infinity. Hence the range of y is [0,∞).

2) √(tanx - tan2x) is defined when tanx - tan2x ≧ 0, i.e.,
0 ≦tanx ≦1
Hence the function is defined for nπ ≦ x < nπ +π/4, n is any integer.
The domain is {[nπ,nπ +π/4] : n is any integer}
For x in this domain, 0 ≦tanx ≦1, 0 ≦ tanx - tan2x ≦ 1/4, hence the range of y is
[0,1/2].