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Find the domain and the range of each of the following real function: f(x) = \(\frac{1}{2-sin3x}\) |
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Answer» Given: f(x) = \(\frac{1}{2-sin3x}\) Need to find: Where the functions are defined. The maximum value of an angle is 2π So, the maximum value of x = 2π/3. Whereas, the minimum value of x is 0 Therefore, the domain of the function, Df(x) = (0, 2π/3). Now, the minimum value of sinθ = 0 and the maximum value of sinθ = 1. So, the minimum value of the denominator is 1, and the maximum value of the denominator is 2. Therefore, the range of the function, Rf(x) = (1/2, 1). |
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