Composition of functions?

Question

kymber · Answer

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parthkohli · Answer

Yes ma'am.

parthkohli · Answer

I think that this is pretty easy for you. We know,$$(g \circ f)(x) = g(f(x)) $$Therefore, our composition simplifies as the following:$$(g \circ f)(x) \implies g(f(x)) \implies g(2x - 1) \implies (2x  - 1) +2 $$

parthkohli · Answer

Do you get the above?

parthkohli · Answer

... or should I be more informal? lol

kymber · Answer

I don't understand what composition means :I

parthkohli · Answer

Never mind that "composition" word. lol

parthkohli · Answer

Do you still understand my explanation?

kymber · Answer

No. I don't understand what I'm supposed to be doing at all! :'(

parthkohli · Answer

I'd give you an example.$$\sqrt{x + 1} $$We're doing two things to $x$. First, we are finding the square root of x, then we are adding 1 to x. So,$$x + 1$$is the composition of two functions: $\sqrt{\text{stuff}}$ and $\text{stuff} + 1$.

parthkohli · Answer

Here, we are again doing two things to $x$.
First, we are multiplying two to it and adding 1(which is $2x - 1$)
Then, we are adding two to that number.

parthkohli · Answer

Operation Multiply Two And Subtract 1 got a boring name, which is f(x).
Operation Add Two To Whatever Comes In Our Way got the name g(x).

parthkohli · Answer

$x$ went to Operation Multiply Two And Subtract 1, and then became $2x - 1$. Sad :(

parthkohli · Answer

And then it went to Operation Add Two To Whatever Comes In Our Way and became $2x - 1 + 2 = 2x + 1$