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

Domain of f(x) -->

Domain of sin(x) is all the Real numbers, `x in R` . For all the real numbers, sine would be in the following range,

`-1lt=sin(x)lt=+1`

Therefore, multiplying by -1,

`+1=gt-sin(x)=gt-1`

Adding +1,

`1+1=gt1-sin(x)=gt1-1`

Therefore,

`2=gt1-sin(x)=gt0`

We know for all real numbers `1- sin(x)`...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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

Domain of f(x) -->

Domain of sin(x) is all the Real numbers, `x in R` . For all the real numbers, sine would be in the following range,

`-1lt=sin(x)lt=+1`

Therefore, multiplying by -1,

`+1=gt-sin(x)=gt-1`

Adding +1,

`1+1=gt1-sin(x)=gt1-1`

Therefore,

`2=gt1-sin(x)=gt0`

We know for all real numbers `1- sin(x)` will be positive.

Therefore the expression inside the squareroot is always positive for all `x in R` . **Therefore the Domain of f(x) is,**

**Domain of** f`(x) = {x in R}`

`f'(x) = (1/2)(-1)(cos(x))(1-sin(x))^(-1/2)`

`f'(x) = (-cos(x))/(2sqrt(1-sin(x)))`

Now since we found the domain and range of 1-sin(x) above, finding the domain of this easy.

The only concern here is the denominator. The denominator cannot be zero. That means,

`1-sin(x) != 0`

`sin(x) != 1`

x not equal to **`npi+(-1)^n(pi/2)` ** where n is any integer.

so x` != {1/2pi, 5/2pi and 9/2pi...}`

or `x!= {1/2pi, 2pi+1/2pi and 4pi+1/2pi...}`

**Therefore domain of f'(x) is given by,**

**Domain of** `f'(x) = {x in R, x != 2npi+1/2pi, n in Z}`

To find the line tangent to f(x) at x =0,

we have to find f'(x) at x = 0.

`f'(0) = (-cos(0))/(2sqrt(1-sin(0)))`

`f'(0) = (-1)/(2sqrt(1-0))`

`f'(0) = -1/2`

This is the gradient of the line,

and we need f(0),

`f(0) = sqrt(1-sin(0))`

`f(0) = sqrt(1-0)`

`f(0) = sqrt(1-sin(x))`

`f(0) = 1`

This is the intercept of the line,

**Therefore the equation is,**

`y = mx+c`

`y = -1/2x+1`