Inverse Functions

For part (a), in the solution they restrict the domain of sin(x^2) to {0,sqrt(pi/2)}. Why did they do this since the image doesn’t contain all possible y values? Does it matter how you restrict the domain then as long as its injective? Is my answer also correct? Bit confused, thanks.

https://imgur.com/a/JhgvvF0

The only way that sin(x^2) can have an inverse is if the domain is restricted so that only one x value pairs with one y value. sinx^2 is a only one to one function in the interval given, and since only one to one functions can have inverses thats why it excludes some y values

Hope this helps (I’m an a level student so I might be wrong)

Yeah I understand that but I thought we had to restrict the domain such that it's one to one and includes all y values. My domain includes all y values whilst still being one to one.

It's like y=sin(x). We restrict the domain from -pi/2 to pi/2 as it's one to one and contains all y values. But here it's like restricting y=sin(x) from 0 to pi/2 which is one to one but doesn't include all y values. This is basically what I'm confused about and whether my answer is still correct or not.

Go on Desmos and plot these. You should see that your answer is not correct since your inverse function is not a reflection of f(x) in the line y=x.

But the given answer is correct.

Since your domain is different, your inverse should be something slightly different. Plot sqrt(pi - arcsin x) against your f(x) ... see if you can figure out whats happening here