We have the function

`f(x) = 3 -2x `

This is a function of the variable `x ` where `x ` can take the value of any number - a small number, a large number, a small negative number or a large negative number. To be formal we might say that the *domain of `x ` *ranges from negative infinity to positive infinity. That is the *domain of `x ` *is `(-infty,infty) `. This unlimited domain is called the *real line *and is what you would normally think of as the `x `-axis or the horizontal axis on a graph, stretching away without limit to the left (negative infinity = `-infty `) and to the right (positive infinity = `+infty `).

Now, the *image* of the function `f ` is the action it carries out on the variable it acts on. It is a mapping of a variable, such as the variable `x `. But any other could be put into the function `f `. It doesn't have to be the variable `x `. In this case the function `f ` multiplies the variable by (-2) and adds 3 to that value.

The question asks what would happen to the variable `2-x ` if we put that into the function `f ` instead of the variable `x `. To make it easier to see what happens here, we can call `2-x ` by another name, a single symbol name, like `y `. So let's write

`2-x = y `

We can now think of `y ` as being a variable much like `x `. In fact, we can just exchange one name for the other - exchange `x ` with `y `.

Because we had `f(x) = 3-2x ` we know just by exchanging/swapping `x ` and `y ` that

`f(y) = 3-2y `

The trick now is to convert this equation in the variable `y ` back to something in terms of `x ` only. We can substitute the value of `y ` in terms of `x ` , that is `y = 2-x `, into `f(y) ` . This gives

`f(y) = 3 - 2(2-x) `

which is the same as writing `f(2-x) = 3 - 2(2-x) `.

It is now a good idea to simplify, or tidy up, this equation to try to get it into the basic form `ax + b ` .

First, multiply out the bracket in the `2(2-x) ` term:

`2(2-x) = 4 - 2x `

This term is *subtracted *from `3 ` so that

`f(y) = 3 - (4-2x) ` which means that

`f(y) = 3 - 4 + 2x ` (remembering a minus times a minus equals a plus).

Collecting the constant terms,

`f(y) = -1 + 2x ` which is equivalent to (swapping the order of terms)

`f(y) = 2x - 1 `

Therefore `f(2-x) = 2x - 1 `

If `y = 2-x ` is put into the function `f ` , the resultant effect (or mapping) is that `x ` is multiplied by 2 and 1 is subtracted from the result.

**If f(x) = 3 - 2x this implies that f(2-x) = 2x - 1. We can work this through by considering a new variable y where we define y = 2-x**

f(2-x)=3-2(2-x) is how you must set up the problem. Simply substitute 2-x for x.

By expanding the equation, you will get f(2-x)=3-4+2x=2x-1.