SOLUTION: Hi all, Im having trouble with the following Function and Domain problem. I have already asked and been given responses and answers for questions a) and b) but was hoping to get so

Question 210295: Hi all, Im having trouble with the following Function and Domain problem. I have already asked and been given responses and answers for questions a) and b) but was hoping to get some help with the c) questions. (See below)
The function f is defined by y = f(x) = 2 ln 4x, 0:01 <= x >= 1.
(a) Solve for x in terms of y, and hence and the formula for the inverse
function f^(-1)(x). Answer: y = (1/4)e^(x/2)
b) Write down the domain of f^(-1) Answer: Since the Range of y = 2*ln(4x) is f(0.01) which is -6.438 < y < 2.7726 the Domain of f^-1(x) is -6.438 < x <= 2.7726
c) Plot f from x = 0:01 to x = 1 and then plot f^(-1) on the same axes but only for domain values of x given by the range of f.
Help with c would be most appreciated.
-Nick.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The function f is defined by y = f(x) = 2 ln 4x, 0:01 <= x >= 1.
c) Plot f from x = 0:01 to x = 1 and then plot f^(-1) on the same axes but only for domain values of x given by the range of f.
-----------------------------------------------
Find the equation of f^-1(x)
Start with y = 2ln(4x)
x = 2ln(4y)
ln(4y) = x/2
4y = e^(x/2)
y = (1/4)e^(x/2)
------------------------
Graph y = 2ln(4x) and (1/4)e^(x/2)
-----

-------------------------
You will have to pick the segment of f(x) where 0.01 and the segment of f^-1(x) that has the range of f as
its domain.
=========================================================
Notice that the two curves are symmetric to the line y = x.
Cheers,
Stan H.
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