Question

How do you show that `f(x)=7x+1` and `g(x)=(x-1)/7` are inverse functions algebraically and graphically?

Answer 1

They are.

Explanation:

Using the inverse composition rule you know that if (fog)=x and (gof)=x then the functions are inverses.
`(gof)=((7x+1)-1)/7` this will result in (gof)=x
`(fog)=7(x-1)/7+1= x-1+1=x`
graphically you can check this if it seems that the functions are reflected over the line x=y.

Answer 2

See below.

Explanation:

For functions:

`f(x) and g(x)`

If `f(x) and g(x)` are inverses, then:

`f(g(x))=g(f(x)=x`

`f(g(x))=7(g(x))+1=7((x-1)/7)+1=(7x-7+7)/7=x`

`g(f(x))=(f(x)-1)/7=(7x+1-1)/7=x`

`:.`

`f(g(x))=g(f(x)=x`

This makes sense if we consider what an inverse function does.

If we put a value `x` into `f(x)` we obtain a value `y`, if we then put this value `y` through the inverse function `g(x)`, we would expect to get back the value we put into `f(x)` which was `x`.

Graphically the inverse function is always a reflection of the function in the line `y=x`:

So:

`g(x)` is a reflection of `f(x)` in the line `y=x`

GRAPH: