Replace y with #f(x)#
#f(x) = 1/3e^(x/2)-2#
Substitute #f^-1(x)# for every x in #f(x)#:
#f(f^-1(x)) = 1/3e^(f^-1(x)/2)-2#
One of the properties of function and their inverse is, #f(f^-1(x)) = x#; this allows us to make the left side become x:
#x = 1/3e^(f^-1(x)/2)-2#
Now, we solve the above equation for #f^-1(x)#
Add 2 to both sides:
#x+2 = 1/3e^(f^-1(x)/2)#
Multiply both sides by 3:
#3x+6 = e^(f^-1(x)/2)#
Use the natural logarithm on both sides:
#ln|3x+6| = ln(e^(f^-1(x)/2))#
Because the natural logarithm and the exponential function are inverses the right side reduces to the argument:
#ln|3x+6| = f^-1(x)/2#
Multiply both sides by 2 and flip the equation:
#f^-1(x) = 2ln|3x+6|#
Multiplication by 2 squares the argument of a logarithm and and, because square cannot be negative, this eliminates the absolute value:
#f^-1(x) = ln((3x+6)^2)#
To verify that the above is truly the inverse, one must show that #f(f^-1(x)) =x# and #f^-1(f(x)) = x#
Start with showing that #f(f^-1(x)) =x#
#f(x) =1/3e^(x/2)-2#
Substitute #f^-1(x) = ln((3x+6)^2)# into the equation:
#f(f^-1(x)) =1/3e^(ln((3x+6)^2)/2)-2#
#f(f^-1(x)) =1/3e^(ln(3x+6))-2#
#f(f^-1(x)) =1/3(3x+6)-2#
#f(f^-1(x)) =x+2-2#
#f(f^-1(x)) =x#
Finish with showing that #f^-1(f(x)) = x#
#f^-1(x) = ln((3x+6)^2)#
Substitute #f(x) = 1/3e^(x/2)-2# for x:
#f^-1(f(x)) = ln((3(1/3e^(x/2)-2)+6)^2)#
#f^-1(f(x)) = ln((e^(x/2)-6+6)^2)#
#f^-1(f(x)) = ln((e^(x/2))^2)#
#f^-1(f(x)) = ln(e^x)#
#f^-1(f(x)) = x#
We have checked both cases and both are true.
#f^-1(x) = ln((3x+6)^2)#