kymber
  • kymber
Composition of functions?
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

kymber
  • kymber
1 Attachment
ParthKohli
  • ParthKohli
Yes ma'am.
ParthKohli
  • 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:\[(g \circ f)(x) \implies g(f(x)) \implies g(2x - 1) \implies (2x - 1) +2 \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

ParthKohli
  • ParthKohli
Do you get the above?
ParthKohli
  • ParthKohli
... or should I be more informal? lol
kymber
  • kymber
I don't understand what composition means :I
ParthKohli
  • ParthKohli
Never mind that "composition" word. lol
ParthKohli
  • ParthKohli
Do you still understand my explanation?
kymber
  • kymber
No. I don't understand what I'm supposed to be doing at all! :'(
ParthKohli
  • 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 \(\text{stuff} + 1\).
ParthKohli
  • 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.
ParthKohli
  • 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).
ParthKohli
  • ParthKohli
\(x\) went to Operation Multiply Two And Subtract 1, and then became \(2x - 1\). Sad :(
ParthKohli
  • ParthKohli
And then it went to Operation Add Two To Whatever Comes In Our Way and became \(2x - 1 + 2 = 2x + 1\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.