How do you find the domain and range of inverse cos(e^x)?

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Answer 1 · 2017-10-06T14:23:13

Domain: #{x in RR : -1<=x<=1}#

Range: #{x in RR : -oo<= y < e^pi}#

#f(x) = cos(e^x)# or #y=cos(e^x)#

We need to make #x# the subject of #y=cos(e^x)#

#cos^-1(y) = cos^-1(cos(e^x))=> cos^-1(y)= e^x#

Taking natural logs of both sides:

#ln(cos^-1(y))= xln(e)#

Dividing by #lne#

#ln(cos^-1(y))/lne= (xln(e))/lne=> ln(cos^-1(y))/lne=x#

#lne=1# ( logarithm of the base is always 1)

So we have:

#x=ln(cos^-1(y))=> f^-1(x) =ln(cos^-1(x))#

#y = ln(cos^-1(x))#

#Cos^-1(x) >= 0=> -1<=x<=1#

So domain is #{x in RR : -1<=x<=1}#

Range:

Maximum value of #cos^-1(x) = pi#, when #x=-1#

#ln(pi) #

Minimum value of #ln(cos^-1(x)) -> -oo#, when #x->1#

#{x in RR : -oo<= y < lnpi}#