Question #bfe8f
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"How can molarity be used as a conversion factor?"
Given: #f(x) = sin^-1( 7x -15)#
Substitute #f^-1(x)# for every #x#:
#f(f^-1(x)) = sin^-1( 7f^-1(x) -15)#
The left side becomes #x# by definition:
#x = sin^-1( 7f^-1(x) -15)#
Take the sine of both sides:
#sin(x) = 7f^-1(x) -15#
Add 15 to both sides:
#sin(x)+15 = 7f^-1(x)#
Divide both sides by 7:
#f^-1(x)= (sin(x)+15)/7#
Before one can declare this as the inverse, one must show that #f(f^-1(x)) = x# and #f^-1(f(x)) = x#:
#f(f^-1(x)) = sin^-1( 7((sin(x)+15)/7) -15)#
#f(f^-1(x)) = sin^-1( sin(x)+15 -15)#
#f(f^-1(x)) = sin^-1( sin(x))#
#f(f^-1(x)) = x#
#f^-1(f(x)) = (sin(sin^-1( 7x -15))+15)/7#
#f^-1(f(x)) = ( 7x -15+15)/7#
#f^-1(f(x)) = (7x)/7#
#f^-1(f(x)) = x#
Q.E.D.
#f^-1(x)= (sin(x)+15)/7#
#f^-1(x)=1/sin^-1(7x-15)#
When #f(x)# is equil to the equation and you add it to the power of -1 then you actually just divide 1 by #f(x)#, thus meaning you should devide 1 with the equation as well. Another answer that would also be correct is just to take the entire equation to the power of -1
#f^-1(x)=(sin^-1(7x-15))^-1# wich should give you the same answer.
