Ugly Inverse Function Graphs & Domain an Range (1 Viewer)

[micuzzo]

hi all... can someone please tell me how i would go about doing these Qs:

1) State the domain and range of y=arcsin x^2 y=x arccos x y=arcsin [tanx]

2) Sketch y= arctan x + arctan (1/x)

Thanks for your help

 

[Drongoski]

hi all... can someone please tell me how i would go about doing these Qs:

1) State the domain and range of y=arcsin x^2 y=x arccos x y=arcsin [tanx]

2) Sketch y= arctan x + arctan (1/x)

Thanks for your help

If I'm not mistaken:

For (2): arctan x + arctan(1/x) = theta + (pi/2 - theta) = pi/2

So it's graph is simply the straight line // to the x=axis: y = pi/2

Looks like someone is pulling our knee.

 

[lolokay]

1) arcsin[x²]
the domain is -1 <= x <= 1
range is 0 <= y <= pi/2, since x² >= 0

x arccos[x]
-1 <= x <= 1
-pi <= y <= ? (i don't know if upper limit is solvable)

y = acrsin[tanx]
-1 <= tan[x] < 1
-pi/4 <= x <= pi/4
-pi/2 <= y <= pi/2

2) (keeping in mind the range of arctan)
let x = tan, -pi/2 < u < pi/2
arctan[tan]
= u
1/x = cot
cot = tan[pi/2 - u] = tan[-pi/2 - u]
arctan[tan[pi/2 - u]]
= pi/2 - u, if 0<= u < pi/2, x >= 0
or arctan[tan[-pi/2 -u]
= -pi/2 - u, if -pi/2 < u < 0, x<0

adding the two we get
actan[x] + arctan[1/x] = u + pi/2 - u
= pi/2 (x >=0)
actan[x] + arctan[1/x] = u - pi/2 - u
= -pi/2 (x < 0)