Find the formula for the inverse function

[fazal]

Homework Statement

The function f is defined by y=f(x) = 3ln4x 0.01<=x<=1

a)Solve for x in terms of y and hence find the formula for the inverse function f^-1(x)

b)Write down the domain of f^-1

c)Plot f from x =0.01 to x=1 and than plot f^-1 on the same axes but only for domain values of x given by the range of f

d)Describe the geometric relationship that you can see between f and f^-1

plse assist as iam very confused with functions...!

Homework Equations

The Attempt at a Solution

 

[Hootenanny]

The Attempt at a Solution

You seem to have missed out an important section.

 

[fazal]

a)Solve for x in terms of y and hence find the formula for the inverse function f^-1(x)

The rule for the inverse function is given by f^-1(x)
how to make y the subject.

b)Write down the domain of f^-1

I suggest that i a graph of y = f(x) over and use it to find the range of f ?

c)Plot f from x =0.01 to x=1 and than plot f^-1 on the same axes but only for domain values of x given by the range of f

d)Describe the geometric relationship that you can see between f and f^-1
are they reflections of each other in the line y = x.??

 

[fazal]

plse assist to answer

 

[Hootenanny]

Okay for part (a) we have:

y = 3ln(4x)

Can you solve for x?

 

[fazal]

for a)
is it this way:


x=1/4e^(y/3)

 

[fazal]

plse assist is this right?

 

[Hootenanny]

for a)
is it this way:


x=1/4e^(y/3)

plse assist is this right?

Yes, that is correct. So now you can write the inverse function as

f^-1(x) = ¹/₄exp(^x/₃)

So, what about the domain of the inverse function?

 

[fazal]

is the domain from 0.2508<=x<=0.3489

by using 1/4 * e^(0.01/3) and 1/4 * e^(1/3)

plse assist to check

 

[Hootenanny]

is the domain from 0.2508<=x<=0.3489

by using 1/4 * e^(0.01/3) and 1/4 * e^(1/3)

plse assist to check

No, that is not correct.

HINT: Think about the range of f(x).

 

[fazal]

yes the range is from 0.01<=x<=1
therefore i plug into the f^-1 and got

0.2508<=x<=0.3489

 

[Hootenanny]

yes the range is from 0.01<=x<=1
therefore i plug into the f^-1 and got

0.2508<=x<=0.3489

No, that is the domain for f(x) and why would you want to plug it into f^-1(x)?

 

[fazal]

i can't seem to figure out the domain...plse assist

b)Write down the domain of f^-1