Question

The range of function f(x) =1/(2-sin3x) is equal to

Answer 1

f(x)= 1/[2 - sin(3x)]  

I will suppose so. The sine function is defined over all real numbers, so the only remaining concern would be division by zero.  

-1 ≤ sin(3x) ≤ 1  

-1 ≤ -sin(3x) ≤ 1  

1 ≤ 2 - sin(3x) ≤ 3  

The denominator is greater than zero for all real x, so the domain of the function is all real numbers.  

1 ≤ 2 - sin(3x) ≤ 3  

1/3 ≤ 1/[2 - sin(3x)] ≤ 1  

1/3 ≤ f(x) ≤ 1  

Domain of f: (-∞, ∞)  

Range of f: [1/3, 1]

Hope this helps!

Answer 2

Answer:

please refer to the attachment

I hope it would help you

thank you.