Free response question for AP calc.: f = sqrt[1 - sinx]

[puppitoe]

My AP Calculus teacher gave us a free response assignment, and I'm slightly confused.

For the following, use f(x) = Squareroot of (1 - sinx)

a) What is the domain of f?

Since you can't take the square root of a negative number, I set 1 - sinx > 0:

. . .1 - sinx > 0
. . .-sinx > -1
. . .sinx < 1
. . .x < sin^-1 (1)
. . .x < 90°

b) Find f'(x)

. . .f(x) = (1 - sin1)^(1/2)
. . .f'(x) = (1/2)(1 - sinx)^(-1/2)(-cosx)
. . .f'(x) = -cosx / [2(1-sinx)^(1/2)]

She doesn't require us to simplify any further.

c) What is the domain of f'(x)

I'm not sure on this one:

. . .1 - sinx > 0
. . .-sinx > -1
. . .sinx < 1
. . .x < sin^-1(1)
. . .x < 90°

d) Write an equation for the line tangent to the graph of f at x = 0

. . .f'(0)= -cos(0) / (2[1-sin(0)]^(1/2))
. . .f'(0) = -1 / 2

But I'm not sure what to do after that...?

If someone could check what I've done, and possibly tell me how to continue and if I've made any mistakes, I'd appreciate it.

 

[skeeter]

My AP Calculus teacher gave us a free response assignment, and I'm slightly confused.

For the following, use f(x) = Squareroot of (1 - sinx)

a) What is the domain of f?

Since you can't take the square root of a negative number, I set 1 - sinx > 0:
isn't sinx < 1 for all x? so, 1 - sinx is greater or equal to 0 for all x ... now, what's the domain?

. . .1 - sinx > 0
. . .-sinx > -1
. . .sinx < 1
. . .x < sin^-1 (1)
. . .x < 90°

b) Find f'(x)

. . .f(x) = (1 - sinx)^(1/2)
. . .f'(x) = (1/2)(1 - sinx)^(-1/2)(-cosx)
. . .f'(x) = -cosx / [2(1-sinx)^(1/2)]

f'(x) is ok

She doesn't require us to simplify any further.

c) What is the domain of f'(x)

1-sinx cannot be 0 ... that means sinx cannot = 1 ... x cannot equal pi/2 ... btw, you need to start thinking in terms of radians rather than degrees

I'm not sure on this one:

. . .1 - sinx > 0
. . .-sinx > -1
. . .sinx < 1
. . .x < sin^-1(1)
. . .x < 90°

d) Write an equation for the line tangent to the graph of f at x = 0

. . .f'(0)= -cos(0) / (2[1-sin(0)]^(1/2))
. . .f'(0) = -1 / 2

But I'm not sure what to do after that...?

f'(0) is the slope of the tangent line at the point on the curve (0, f(0)) ...
remember how to use the point-slope form of a linear equation?

If someone could check what I've done, and possibly tell me how to continue and if I've made any mistakes, I'd appreciate it.

 

[tkhunny]

I'm still scratching my head over the use of the inverse sine in an inequality. Why do you think you can do that? For one, the sine is a periodic function. You should think about that one a little harder. Not to mention the conscience-clearing conclusion exposed by skeeter.