How do you find the domain and range of # y = sin^(-1)(x^2)#?

1 Answer
Dec 16, 2017

Answer:

Domain: #{x|-1<=x<=1}#
Range: #{y|0<=y<=pi/2}#

Explanation:

Domain of the function
We know the inverse sin function is only defined for #-1<=x<=1# (since #sin# only takes on those values), so we know that our function is only defined when #x^2# obeys that same interval. We figure out when this is true by solving the following inequality:
#-1<=x^2<=1#

The first one, #-1<=x^2#, is true for all real numbers, #x in RR#

And we can solve the second one by taking the square root on both sides:
#x^2<=1#

#sqrt(x^2)<=sqrt(1)#

#x<=1#

Now we combine this with our original interval, which we do by taking the most restrictive of the bounds:
#{x|-1<=x<=1}#

And this is when our function is defined, so it must be the domain.

Range of the function
The inverse sine function usually ranges from #-pi/2# to #pi/2#, but since negative values are only produced by negative inputs to the function, the #x^2# makes it so that we will only ever get positive (or #0#) values. This means our range will be:
#{y|0<=y<=pi/2}#