Explain, using appropriate theorems, why the function f(x)=sin^-1(x^2-1)is continuous at every number in its domain. State the domain.

Answer 1

Recall that sin^-1(x^2-1) represents an angle a whose sine is equal to x^2-1, that is sin(a)=x^2-1. In other words x^2-1 is a value of the sine function, therefore it must lie in the interval [-1,1]:

-1<= x^2-1 <= 1, or in other words 0<= x^2 <= 2. Of course x^2 is always non-negative, so the first inequality is always true and we end up getting that the domain of f(x) is given by x^2<=2, or equivalently [-sqrt(2), sqrt(2)].

Regarding continuity: x^2-1 is a polynomial and so is continuous everywhere, sin^-1(y) is continuous in its domain (the interval [-1,1]) as an inverse trigonometric function. Therefore, their composition f(x) is also continuous in its domain [-sqrt(2), sqrt(2)].

Answer 2

Here's how you would start with it. I'll leave a lot of the details of the proof to you, but here are the arguments for it.

Let's look at f(x) = sin-1(x). This is the inverse function of some g(x) which is a subset of sin(x). By that I mean the domain of g(x) is a subset of the real number line, and g(x) = sin(x) for that domain. The problem is that sin(x) is a function on all real x, yet it is not a one-one function. We must restrict g's range if we are to create a one-one function. One way to do that is let g = sin(x) only for -π/2 ≤ x ≤ π/2. Then g is one-one as it passes the horizontal line test as well. Then show that if f(x) = g-1(x), then the domain of f(x) is -1 ≤ x ≤ 1

This means the what goes inside the sin-1( ) function, or the argument must follow that domain as well. Show that if f(x) = sin-1(x^2 - 1), then -√2 ≤ x ≤ √2 (Note: that x = 0 is included in the domain, because of the open inequality).

Now let's show that every x in this domain is continuous in f(x) given. For f(x) to be continuous at x, we require that for some other points in the domain, called s,

lim[f(s)]s→x from the left = lim[f(s)]s→x from the right = f(x)

Note, that if x is a boundary point in the domain, and no values exist to its right (for example x = √2) then we don't need to check anything to its right. f(x) does not blow up to inf or -inf in our domain. Also by graphing it, we can see that there are no discontinuities (there many be another way to show this that I'm missing). This is also true because sin-1(x) is itself a continuous function, and our argument stays within the domain of sin-1(x). Check x = 0, x = √2, x = -√2 to confirm this.

Good luck!