## kymber Group Title Composition of functions? 2 years ago 2 years ago

1. kymber

2. ParthKohli

Yes ma'am.

3. ParthKohli

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:$(g \circ f)(x) \implies g(f(x)) \implies g(2x - 1) \implies (2x - 1) +2$

4. ParthKohli

Do you get the above?

5. ParthKohli

6. kymber

I don't understand what composition means :I

7. ParthKohli

Never mind that "composition" word. lol

8. ParthKohli

Do you still understand my explanation?

9. kymber

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

10. ParthKohli

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 $$\text{stuff} + 1$$.

11. ParthKohli

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.

12. ParthKohli

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).

13. ParthKohli

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

14. ParthKohli

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