Question

Let `f(x)=5x+4` and `g(x)=x−4/5` , find: a). `(f@g)(x)` ? b). `(g@f)(x)` ?

Answer 1

`(f ∘ g) (x) = 5x` `(g ∘ f) (x)=5x+16/5`

Explanation:

Finding `(f ∘ g) (x)` means finding `f(x)` when it is composed with `g(x)`, or `f(g(x))` . This means replacing all instances of `x` in `f(x)=5x+4` with `g(x)=x-4/5`:

`(f ∘ g) (x)=5(g(x))+4=5(x-4/5)+4=5x-4+4=5x`

Thus, `(f ∘ g) (x) = 5x`

Finding `(g ∘ f) (x)` means finding `g(x)` when it is composed with `f(x)`, or `g(f(x)).` This means replacing all instances of `x` in `g(x)=x-4/5` with `f(x)=5x+4:`

`(g ∘ f) (x) = f(x)-4/5=5x+4-4/5=5x+20/5-4/5=5x+16/5`

Thus, `(g ∘ f) (x)=5x+16/5`

Answer 2

See explanation...

Explanation:

Alright, first lets remember what `f@g` and `g@f` mean.

`f@g` is a fancy way of saying `f(g(x))` and `g@f` is a fancy way of saying `g(f(x))`. Once we realize this, these problems aren't that difficult to solve.

So `f(x)=5x+4` and `g(x)=x-4/5`

a) `f@g`

Ok lets start with the `f(x)` function

`f(x)=5x+4`

Then, we just add the `g(x)` function whenever we see an `x` in the `f(x)` function.

`f(g(x))=5g(x)+4` `->` `5(x-4/5)+4`

Simplify:

`f(g(x)))=(5x-4)+4` `->` `5xcancel(-4)cancel(+4)`

So therefore, `f@g=5x`

b) `g@f`

Alright, it's the same process here just it's the opposite. Let's start with the `g(x)` function.

`g(x)=x-4/5`

Then, we just add the `f(x)` function whenever we see an `x` in the `g(x)` function.

`g(f(x))=f(x)-4/5` `->` `(5x+4)-4/5`

Simplify:

`g(f(x))=5x+16/5`

Therefore, `g@f=5x+16/5`

Hope this helped!
~Chandler Dowd

Answer 3

For `g(x)=x-4/5` it is solved by Chandler Dowd and VNVDVI
For `g(x)=(x-4)/5`, [ requested by Widi K. ]the solution is

`color(red)( (fog)(x)=x and (gof)(x)=x)`

Explanation:

We have,`f(x)=color(red)(5x+4 ...to(1)`
`and g(x)=color(blue)((x-4)/5.......to(2)`.
Hence,
`(fog)(x)=f(g(x))`
`(fog)(x)=f(color(blue)((x-4)/5))....to`from(2)
`(fog)(x)=f(m)`,......[ take `m=(x-4)/5` ]
`(fog)(x)=color(red)(5m+4`......[Apply (1) for `x tom`]
`(fog)(x)=cancel5(color(blue)((x-4)/cancel5))+4`...[ put `m=(x-4)/5` ]
`(fog)(x)=x-4+4`
`(fog)(x)=x`

`(gof)(x)=g(f(x))`
`(gof)(x)=g(color(red)(5x+4))......to`from(1)
`(gof)(x)=g(n)........`[ take `n=5x+4`
`(gof)(x)=(color(blue)((n-4)/5))`......[ Apply (2) for `x ton`
`(gof)(x)=(5x+4-4)/5....`[ put `n=5x+4 ]`
`(gof)(x)=(5x)/5`
`(gof)(x)=x`