Finally though I understood composite functions... then

[GetThroughDiffEq]

Okay, so I finally thought that I understood composite functions last evening:

[f o g](x) is simply the equation of g(x) with its x replaced with whatever the value of f(x) is.

Then, I got a little confused when it came to fraction (see example):

Say we are to determine the values of both [f o g](x) and [g o f](x), given that f(x)=20x-2/5 and g(x)=x-9

*Now, to solve [f o g](x), is the portion of x isolated from -9?

[E.G. (20x-2/5 <<< Solve by dividing first before subtracting the 9) -9]

OR

Would it just be (20x-2-9)/5?

 

[lev888]

[f o g](x) is simply the equation of g(x) with its x replaced with whatever the value of f(x) is.

(f∘g)(x) is f(g(x)). Therefore, it's f(x) with x replaced by g(x).
Don't quite understand your question. Can you post your solutions for both cases?

 

[JeffM]

Okay, so I finally thought that I understood composite functions last evening:

[f o g](x) is simply the equation of g(x) with its x replaced with whatever the value of f(x) is.

Then, I got a little confused when it came to fraction (see example):

Say we are to determine the values of both [f o g](x) and [g o f](x), given that f(x)=20x-2/5 and g(x)=x-9

*Now, to solve [f o g](x), is the portion of x isolated from -9?

[E.G. (20x-2/5 <<< Solve by dividing first before subtracting the 9) -9]

OR

Would it just be (20x-2-9)/5?

Is

[MATH]f(x) = 20x - \dfrac{2}{5}[/MATH]
or is it

[MATH]f(x) = \dfrac{20x - 2}{5}[/MATH]
The idea of a composite function is one that is so simple that students drive themselves crazy trying to find some weird complexity. The idea is that you put a number into a function, a number pops out, and then you put that number into another function. That's it.

[MATH]\text {If } \alpha(2) = 17 \text { and } \beta (17) = 30 \text{, then } \beta ( \alpha (2))) = \beta (17) = 30.[/MATH]
Do you remember PEMDAS from the first weeks of algebra? You work parentheses from the inside out.

But if we want to find a formula for such a composite, you can think it through by adding variables.

[MATH]f(x) = \dfrac{2x -2}{5} = u.[/MATH]
[MATH]g(x) = x - 9 = v.[/MATH]
[MATH]\therefore f(g(x)) = f(v) = \dfrac{2v - 2}{5} = \dfrac{2(x - 9) - 2}{5} = \dfrac{2x - 20}{5}.[/MATH]
Let's check.

[MATH]g(25) = 25 - 9 = 16.[/MATH]
[MATH]f(16) = \dfrac{2 * 16 - 2}{5} = \dfrac{30}{5} = 6.[/MATH]
[MATH]\therefore f(g(25)) = 6.[/MATH]
What does the formula give us?

[MATH]\dfrac{2 * 25 - 20}{5} = \dfrac{30}{5} = 6,[/MATH]
which is the correct answer.