Question

f(x)=arccos(1-X^2)

What is the domain of f?

Answer 1

@Will.H

Answer 2

Keep in mind that "arccos x" can be read as "the angle whose cosine is x (or x/1)."
When we're speaking of the sine and cosine functions, what's the largest value either can have? In other words, what is the smallest value, and what is the largest, that sin x can have?

Answer 3

The only thing I know about arccos it that it´s the inverse to cosine, which means its domain and range should be switched, right?

Answer 4

The smallest is -1 and largest is 1 right?

Answer 5

Yes. If you begin with y=sin x, and come up with the set to which input values x are limited, and the range of the function y=sin x, you can reverse order every point and obtain a listing of points lying on the inverse function y=arctan theta.

If (pi/3, 1/2) is on the graph of y=cos x, then (1/2, pi/3) is on the graph of y=arccos x.

Answer 6

What are the domain and range of y=sin x?

Interchange domain and range here, and you'll end up with the domain and range of y=arcsin x.

Answer 7

Well the domain of sin is (−∞,∞) because it just continues and range is [−1,1] so the opposite is....the opposite, [−1,1] in domain and (−∞,∞) in range.

Answer 8

Things aren't as clean as that for functions that aren't typically invertible. \(f(x)=\cos x\) is not invertible because it's not one-to-one; for one instance, \(\cos0=\cos2\pi=1\). To get around this you need to restrict the usual domain of \(\cos x\) to a branch of it that is one-to-one. A standard choice is the interval \([0,\pi]\).

The restriction of the domain to \(x\in[0,\pi]\) retains the range \(\cos x\in[-1,1]\) and this "partial" cosine function, call it \(\bar{f}(x)\), is one-to-one so an inverse exists. We call this inverse \(\bar{f}^{-1}(x)=\arccos x\).

If the domain and range of \(\bar{f}\) are \([0,\pi]\) and \([-1,1]\), then the domain and range of \(\bar{f}^{-1}\) are \([-1,1]\) and \([0,\pi]\), respectively.

This means that for \(\arccos(1-x^2)\) to be defined, the domain must satisfy \(-1\le1-x^2\le1\). What would that be in terms of \(x\) explicitly?