Question #4f389

1 Answer
Dec 7, 2017

#f^-1(x) = ln((3x+6)^2)#

Explanation:

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)#