Need help with functions

Question

h(x)=1/1+x=0.1, so x=9, whats h^-1(x) and how to find it?

Answer 1

Hi Erlandas;

First, I am just going to work with...

h(x)=1/(1+x)

This can also be represented as...

y=1/(1+x)

To find h^-1(x), we first need to isolate x, the way y is currently isolated...

Let's multiply both sides by (1+x)

y(1+x)=[1/(1+x)](1+x)

y(1+x)=1

Let's divide both sides by y...

[y(1+x)]/y=1/y

1+x=1/y

Let's subtract 1 from both sides...

1+x-1=(1/y)-1

x=(1/y)-1

x is now isolated.  Let's switch x and y.  This is h^-1(x), also known as the inverse function...

y=(1/x)-1

Let's plug-in x=9

y=(1/9)-1

y=-8/9