How do you find the domain of #f(x)=(x^2-4x+4)/(2sin(x))#?

1 Answer
Jan 25, 2017

Answer:

#f(x)=(x^2-4x+4)/(2sinx), x in RR, x!=npi#/#x!=180n#

Explanation:

The domain of a function is all the values of #x# for which the function is defined. Since the function is a quotient, the only values it isn't defined for is when the denominator is equal to zero.

Since the denominator is a #sin# function, we know that it will be equal to zero whenever #x# is a multiple of #pi^"c"# or #180^"o"#. So we can write this as #x!=npi#.