What is the domain and range of #sqrtcos(x^2)#?

1 Answer
Jul 4, 2018

Answer:

#x in ...U [ -sqrt((5pi)/2), -sqrt((3pi)/2) ] U [ -sqrt(pi/2), sqrt(pi/2) ] U [ sqrt((3pi)/2), sqrt((5pi)/2) ] U [ sqrt((7pi)/2), sqrt((9pi)/2) ] U ...# and #y in [ 0, 1 ]#

Explanation:

#0 <= cos ( x^2 ) <= 1#, and so,

#0 <= y = sqrt cos (x^2 ) <= 1#. Inversely,

#x = +- sqrt( abs( cos^(-1)y^2)), +- sqrt(2kpi +- cos^(-1)y^2)#,

#k = 1, 2, 3, ...#, avoiding negatives under the radical sign.

So,

#x in ...U [ -sqrt((9pi)/2), -sqrt((7pi)/2) ] U#

#[ -sqrt((5pi)/2), sqrt((3pi)/2) ] U [ -sqrt(pi/2), sqrt(pi/2) ] #

#U [ sqrt((3pi)/2), sqrt((5pi)/2) ] U [sqrt((7pi)/2), sqrt((9pi)/2) ]U...#

See the phenomenal graph.
graph{y - sqrt(cos(x^2))=0[-10 10 -1 9]}