Question 1203320: Hi,
I need some help with domain and range questions with trig.
a) y=xsin^-1(x^2)
b)y=tan^-1 (sqrt(1-x^2))
I get the same domain for both, being -1 and 1.
Range for a) get -pi/2 and pi/2 and b) 0 and pi/2
I'm not sure if i'm correct and if so not quite sure I understand how I got my answers.
Thank you
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
(a) 
domain...
x^2 is 0 or positive, and the argument for inverse sine must be between -1 and 1. So the domain for  is from -1 to 1; and since x is between 0 and 1, the domain for , is still from -1 to 1.
ANSWER: -1 to 1 (your answer is correct)
range...
has its maximum value when x^2=1, which occurs at both -1 and 1, the boundaries of the domain. At x=-1 or x=1, where x^2 is 1, inverse sine is pi/2. So at x=-1 the value of is -pi/2, and at x=1 the value is pi/2.
ANSWER: -pi/2 to pi/2 (your answer is correct)
(b) 
domain...
The argument for square root must be non-negative, so again x must be between -1 and 1.
ANSWER: -1 to 1 (your answer is correct)
range...
At both x=-1 and x=1, 1-x^2 is 0, and inverse tangent of 0 is 0.
Between x=-1 and x=1,  is positive, with a maximum of 1 when x is 0. Inverse tangent of 1 is pi/4.
ANSWER: 0 to pi/4 (your answer was 0 to pi/2)
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