# How do you find the implied range and domain of arccos((x-1)^2)?

Jul 7, 2018

Range is $\left[0 , \frac{\pi}{2}\right]$. Domain for x is $\left[0 , 2\right]$. Also introduced is the inverse operator ${\left(\cos\right)}^{- 1}$, on par with ${f}^{- 1}$.

#### Explanation:

Over centuries, we have been told that the range of ${\cos}^{- 1} x$

or, for that matter, $\arccos x$ is $\left[0 , \pi\right]$.

In the ${f}^{- 1}$ sense, I like to use ${\left(\cos\right)}^{- 1}$ to state that

the range of ${\left(\cos\right)}^{- 1} x$ is $\left(- \infty , \infty\right)$.

Here,

the conventional range of $y = \arccos \left({\left(x - 1\right)}^{2}\right)$ is $\left[0 , \pi\right]$

and , as $\cos y \ge 0$, the effective range is $\left[0 , \frac{\pi}{2}\right]$.

The range of $Y = {\left(\cos\right)}^{- 1} {\left(x - 1\right)}^{2}$ is union of

$\left\{\left[\begin{matrix}2 k - \frac{1}{2} \pi \\ 2 k + \frac{1}{2} \pi\end{matrix}\right]\right\} , k = 0 , \pm 1 , \pm 2 , \pm 3 , . .$.

Note that, piecewise,

$Y = {\left(\cos\right)}^{- 1} {\left(x - 1\right)}^{2} = 2 k \pi \pm \arccos \left({\left(x - 1\right)}^{2}\right)$,

$k = 0 \pm 1 , \pm 2 , \pm 3 , . .$, for Y in the respective period

$\left[2 k \pi , \left(2 k + 1\right) \pi\right]$.

Cosine value $\in \left[- 1 , 1\right]$. Yet, here,

$0 \le {\left(x - 1\right)}^{2} = \cos y \in \left[0 , 1\right]$. So,

$\left\mid x - 1 \right\mid \le 1$. And so, x-domain is given by

$0 \le x \le 2$.

y-graph:
graph{(y - arccos ((x-1)^2))((x-1)^2+(y-pi/2)^2-.04)=0}
Note the crest at $\left(1 , \frac{\pi}{2}\right)$

Y-graph:
graph{cos y - (x-1 )^2 =0 [0 60 -15 15]}
Graph for understanding Y-range:
graph{cos y-(x-1)^2=0[0 4 -10 10]}

This is the double graph for $x - 1 = \pm \sqrt{\cos y}$
The two separate graphs in the combined Y-graph:

graph{(cos y)^0.5 - (x-1 )=0 [0 60 -15 15]}
graph{(cos y)^0.5 +(x-1 ) =0 [0 60 -15 15]}

You can realize now the suppressed details in our restricted range

$\left[0 , \pi\right]$.