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

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Answer 1 · 2018-07-07T05:48:32

Range is # [ 0, pi/2 ]#. Domain for x is #[ 0, 2 ]#. Also introduced is the inverse operator #(cos)^(-1)#, on par with #f^(-1)#.

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]#.