what is the domain and range for sqrt(1-2sin(x))?

Question

assume we do not know how the graph looks like. Please show working out and explanation. no calc and graphic calc can be used.

Answer 1

f(x) = sqrt(1 - 2sin(x))

To find the domain, the restriction is that the inside of the square root is greater than or equal to 0. That is,

1 - 2sin(x) >= 0

Solving this inequality,

-2sin(x) >= -1
sin(x) <= 1/2

From the interval 0 to 2pi, sin(x) is less than or equal to 1/2 at the following intervals: {this is using the knowledge of our unit circle}

[0, pi/6] U [5pi/6, 2pi)

Without the interval restriction, this is from 5pi/6 to 13pi/6, adding 2k(pi) to each.

Domain: [5pi/6 + 2k(pi) , 13pi/6 + 2k(pi)]

To obtain the range, start with the range of sin(x) and perform operations to get it to look like f(x).

-1 <= sin(x) <= 1
Multiply everything by -2,

2 >= -2sin(x) >= -2
Add 1,

3 >= 1 - 2sin(x) >= -1
Take the square root;

sqrt(3) >= sqrt(1 - 2sin(x)) >= sqrt(-1)
Remove the sqrt(-1) and replace it with 0; this is using our knowledge that the sqrt function always produces a positive value.

sqrt(3) >= sqrt(1 - 2sin(x)) >= 0

Therefore, our range is [0, sqrt(3)]

Answer 2

Since negative square roots are tacky, -2 sin(x) should not exceed 1 or sin(x) exceed 0.5. The domain is then from - 7 pi/6 to pi/6. (From -pi to 0, sin(x) is negative). Range is from 0 to sqrt(3).

Answer 3

domain: since if sin(x)>0.5 eqn will not make sense,
x_<(pi)/3 or x>_5(pi)/3
range: since -1_ x is in radians, or for degrees, x_<30 or x>_150