How do you find the domain and range of y = sin^(-1)(x^2)y=sin−1(x2)?

1 Answer
Alvin L.
Dec 16, 2017

Domain: {x|-1<=x<=1}{x∣−1≤x≤1}
Range: {y|0<=y<=pi/2}{y∣0≤y≤π2}

Explanation:

Domain of the function
We know the inverse sin function is only defined for -1<=x<=1−1≤x≤1 (since sinsin only takes on those values), so we know that our function is only defined when x^2x2 obeys that same interval. We figure out when this is true by solving the following inequality:
-1<=x^2<=1−1≤x2≤1

The first one, -1<=x^2−1≤x2, 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}

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