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

1 Answer
Jul 7, 2018

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].