What is the range of f(x)= 1/(1-sin(x)) ?

The range of f(x)= 1/(1-sin(x)) is f(x)>= 1/2

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range of f(x)= 1/(1-sin(x))

range

f (x) = \frac{1}{1 - sin ( x )}

f (x) \geq \frac{1}{2}

Interval Notation

[\frac{1}{2}, \infty)

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

\frac{1}{1 - sin ( x )} : 2 πn \leq x < \frac{π}{2} + 2 πn or \frac{π}{2} + 2 πn < x < 2 π + 2 πn

Extreme Points of

\frac{1}{1 - sin ( x )} :

Minimum

(\frac{3 π}{2} + 2 πn, \frac{1}{2})

Find the range for the interval

0 \leq x < \frac{π}{2} : \frac{1}{2} \leq f (x) < \infty

Find the range for the interval

\frac{π}{2} < x < 2 π : \frac{1}{2} \leq f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

f (x) \geq \frac{1}{2} or f (x) \geq \frac{1}{2}

f (x) \geq \frac{1}{2}

f (x) = (2 x - 8)^{\frac{2}{3}}

x^{3} - \frac{3}{2} x^{2}, - 5 \leq x \leq 4

f (x) = \frac{x ^{2} + 1}{2 x ^{2} + 7}

4 (\frac{1}{2})^{x - 3} - 2

y = 5 x^{2}

What is the range of f(x)= 1/(1-sin(x)) ?
The range of f(x)= 1/(1-sin(x)) is f(x)>= 1/2