Question

Let `f(x)=5x-4 and g(x)=x+7`. Determine `color(white)("d") a: (f@g)(x), color(white)("d")b: (g@f)(x),color(white)("d")c: (f@g)(3)color(white)(...)` ?

this is intermediate algebra and I don't understand how to solve this problem regarding functions

Answer 1

A lot of explanation given.

Part a: `(f@g)(x)=z=5x+31`

Part b: `(g@f)(x)=5x+3`

Part c: `(f@g)(3)=15+31`

Explanation:

`color(magenta)("Explaining what it all means")`

The letters `f and g` are a sorts of names that the writer may assign to any particular expression structure they so choose.

For example, if I use the word 'wheel' you immediately know what I am talking about it and possibly a picture of a wheel occurs in your mind.

Equally the original writer of this question intended the structure `f` when applied to 'something' to mean `5xx"something" -4`

So `f` of 'something' `color(white)("ddd")`is written as:

`f("'something'")->5xx"'something'"-4`

so `f(x)->5xx x-4 color(white)("ddd")->color(white)("ddd") 5x-4`

In the same way the different structure name `g` when applied to `x` does this to `xcolor(white)("d")->color(white)("d")x+7` so we have:

`g(x)=x+7`
`g(b)=b+7`
`g("wheel")="wheel"+7` and so on
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We can nest these (put one inside the other) and one possible way of demonstrating this is:

Set `color(red)(y=g(x)=x+7)`

Set `color(blue)(z=f(color(red)(y))=5color(red)(y)-4)`

The mathematical symbols: `(f@g)(x)` is saying take the structure of `g` applied to `x` and then apply the structure of `f` to that. In a way it is a bit like substitution as I have shown above. The form `(f@g)(x)` is worded: `f` of `g` of `x`

Ok so lets do it!
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`color(magenta)("Answering part a: "(f@g)(x))`

`g` is in the right so that is the substitution.

`color(red)(y=g(x)=x+7)`

`color(blue)(z=f(color(red)(y))=5color(red)(y)-4)`

Thus by substituting for `y` we have:

`color(blue)(z=f(color(red)(y))=5color(red)((x+7))-4)`

`z =f(y)=color(white)("d") 5x+35-4`

`(f@g)(x)=z=5x+31 ......................Equation(1)`
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`color(magenta)("Answering part b: "(g@f)(x))`

`f` is on the right so that comes first and is the substitution

`color(red)(y=f(x)=5x-4)`

`color(blue)(z=g(color(red)(y))= color(red)(y)+ 7`

`color(blue)(z=color(red)(5x-4) +7)`

`z=g(y)=5x+3`

`(g@f)(x)=5x+3 ................Equation(2)`
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`color(magenta)("Answering part c: "(f@g)(3))`

Using `Equation(1)`

`(f@g)(x)=5x+31`

`(f@g)(3)=5(3)+31`

`(f@g)(3)=15+31`

`(f@g)(3)=46`