Find the domain and range of $y=\cos^{-1} \sqrt{1-x}$

Question

Find the domain and range of $y=\cos^{-1}\sqrt{1-x}$.

Can someone please help me with question above, as to how it's done? Thanks.

I am unfamiliar with what you do when there is a square root.

Answer 1

Domain is $$0\leq x\leq 1$$ and range is $$0\leq y \leq \frac{\pi}{2}$$

As $cos^{-1}$x operates in $[-1,1]$ hence the argument must be between these hence the domain $0\leq x\leq 1$ also since domain is $[0,1]$ hence the range $0\leq y \leq \frac{\pi}{2}$.

$**Note**$: If you can determine the domain of a continous function between two intervals then plugging in the extreme values in the function provide you with the range.

Don't know of any exceptions yet.

Answer 2

Well, if we are in the reals, then the inside of the square root must be positive. So $1-x>0$: $x<1$, and the inverse cosine must operate on numbers on the interval $(-1,1)$, so $\sqrt{1-x}$ must also be between $(-1,1)$:

$$ 0 \leq 1-x \leq 1$$

We can deduce

$$x \leq 1 \qquad \text{and} \qquad x\geq0$$

EDIT: I forgot the range: arccosines are multivalued, and you choose by convention the values in between $[0,\pi]$, so that will be the range

Answer 3

The domain of the function $\cos^{-1}$ is $[-1,1]$ and its range is $[0,\pi]$ so the domain $D$ of $\cos^{-1}\sqrt{1-x}$ is $$D=\{x\in \mathbb R\ |\ 0\leq1-x\leq 1\}=[0,1]$$

and if $x\in D$ then $\sqrt{1-x}\in [0,1]$ so the range of the given function is $[0,\frac{\pi}{2}]$.