Let f(x)=(x-3)/x , x>0. Show f is one-to-one and find the inverse function for f.

Question

Calculus 2 homework that I can't seem to figure out for the life of me at the moment.

Answer 1

We want to show the inverse so we need to change our x to y.

Thus,

y = x-3/ x

x = y-3/ y

Step1 : solve for y

x = 1-3/ y

x-1 = -3/ y

y = -(3 / x-1)

Thus, f^-1 = -(3/x-1)

So, if we want to show that it is 1-1. We need to solve for x-1=0.

Thus f has to be x>0. To say our f is 1-1 such that our f^-1 is continuous whenever x does not equal 1.

Answer 2

Firstly, to show that f(x) is 1-1, you can verify the following:

if f(x) = f(y), then x = y. That is, for every output, there is a unique input. So,

1 - 3/x = 1- 3/y implies

3/x = 3/y (now invert both sides)

x/3 = y/3 (finally, multiply both sides by 3)

x = y. Therefore, f is 1-1.

Now, to find the inverse function, we swap independent and dependent variables. Let the inverse function be g(x) So,

x = 1-3/g(x) implies

1-x = 3/g(x) implies

1/(1-x) = g(x)/3 implies

g(x) = 3/(1-x)