A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

f(x) = x + 1, g(x) = √(x − 3) ; What is the domain of f ◦ g? I just want to know if it's [3,infinity) or [2,infinity) because of the restrictions on compositions of functions.

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Usually what i do is compose them , and don't simplify it. then look for the composed domain

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    f ◦ g(x) = f(g(x)) = f ( √(x − 3)) = √(x − 3) + 1 without simplify, the domain is x >= 3

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the reason why i say don't simplify, let me illustrate with an example. suppose you are given f(x) = x^2 , g(x) = √ (x-3) then f(g(x)) = (√(x-3) )^2 if you simplify this you get x-3. The domain of x -3 is all the real numbers. However that is not the domain of f ◦ g , the domain is x >= 3.

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Perfect. The textbook is wrong-- just as I thought. Thanks!

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.