Let f(x)=sqrt(1-sinx) What is the domain of f(x), find f'(x), what is the domain of f'(x) and write an equation for the line tangent to the graph of f at x=0

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`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)`...

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`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`

 

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