Question

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

Answer 1

Range is `[ 0, pi/2 ]`. Domain for x is `[ 0, 2 ]`. Also introduced is the inverse operator `(cos)^(-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 `[ 0, pi ]`.

In the `f^(-1)` sense, I like to use `(cos)^(-1)` to state that

the range of `(cos)^(-1) x` is `( - oo, oo )`.

Here,

the conventional range of `y = arccos( ( x - 1 )^2)` is `[ 0, pi ]`

and , as `cos y >=0`, the effective range is `[ 0, pi/2 ]`.

The range of `Y = (cos)^(-1) ( x - 1 )^2` is union of

`{[ (2k-1/2)pi, (2k+1/2)pi]}, k = 0, +-1, +-2, +-3, ..`.

Note that, piecewise,

`Y = (cos)^(-1) ( x - 1 )^2 = 2kpi +- arccos( ( x - 1 )^2)`,

`k = 0 +- 1, +-2, +- 3, ..`, for Y in the respective period

`[ 2kpi, (2k+1)pi ]`.

Cosine value `in [ - 1, 1 ]`. Yet, here,

`0 <= ( x - 1 )^2 = cos y in [ 0, 1 ]`. So,

`abs ( x - 1 ) <= 1`. And so, x-domain is given by

`0 <= x <= 2`.

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

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 = +-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

`[ 0, pi]`.