Question #259000

Find the domain and range of the function f defined by f(x) = 1/1-sinx

1
Expert's answer
2021-11-04T19:54:24-0400

(a) The domain of a function is the set of input or argument values for which the function is real and defined.

Domain of 11sinx\frac{1}{1-sinx}:



=[solution:2πx<π2+2πn or π2+2πn<x<2π + 2πninterval Notation[2πn, π2+2πn) (π2+2πn, 2π+2πn)]=\begin{bmatrix} solution:& 2\pi \le x \lt \frac{\pi}{2 }+2\pi n\space or\space \frac{\pi}{2 }+2\pi n\lt x\lt 2\pi \space+\space2\pi n \\ interval \space Notation & \lbrack2\pi n,\space\frac{\pi}{2} + 2\pi n\rparen\bigcup \ \lparen\frac{\pi}{2}+2\pi n,\space2\pi+2\pi n\rparen \end{bmatrix}

=2πnx<π2+2πn or π2+2πn<x<2π+2πn=2\pi n \leq x \lt \frac{\pi}{2} +2\pi n \space or \space \frac{\pi}{2} +2\pi n \lt x \lt 2\pi +2\pi n

The range is the set of values of the dependent variable for which a function is defined

Range of 11sinx: =[solution:12f(x)1interval notation:[12,1]]Range \space of\space\frac{1}{1-sinx}:\space= \begin{bmatrix} solution: & \frac{1}{2} \leq f(x)\leq 1 \\ interval\space notation: &\lbrack \frac{1}{2},1\rbrack \end{bmatrix}



Range=[12,1]Range = \lbrack \frac{1}{2}, 1 \rbrack


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