Results 1 to 8 of 8

Math Help - Easy Questions, need a reminder

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    123

    Easy Questions, need a reminder

    Can someone please help me out with these question's. I'm trying to add, subtract, multiply and divide functions.

    How will I do that for:

    f (x) = x^2 - 3 g (x) = 2 - square root (x+1)

    Basically the way i did it was.


    Add: x^2 - 3 + 2 - square root of (x+1)
    = x^2 - square root of (x+1) - 1

    Subtract: x^2 - 3 - 2 + square root (x+1)
    = x^2 + square root (x+1) - 5


    Multiply: I'm not sure how to do this one ... but i got:

    f (x) . g (x) = (x^2-3) (2- square root (x+1)

    = 2x^2 - square root (x+1)^3 - (square root (x+1)^3 + 3 square root (x+1)

    but it's probably wrong.


    Can someone help me with these, thanks.

    Divide: i'll figure it out after i know how to multiply
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    I'm not sure where you got that...

    (x^2-3) (2- \sqrt{x+1})  = 2x^2 - x^2\sqrt{x+1} - 6 + 3\sqrt{x+1}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    123
    Actually for the 3rd part: multiplication I got:

    (x^2-3) X (2- square root (x+1)

    = 2x^2 - square root (x+1)^3 - 6 + 3 square root (x+1)

    = square root (x+1)^3 + 2x^2 + 3 square root (x+1) - 6


    Thanks I think I sort of get it. Did I do the adding/subtracting right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    x^2 \times \sqrt{x+1} = x^2\sqrt{x+1} = \sqrt{x^4(x+1)} = \sqrt{x^5 + x^4}

    I don't know how you got \sqrt{(x+1)^3}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    Posts
    123
    Thanks, I made an error when I tried to multiply it... I havne't doen math in a while and forget a lot of the more simpler stuff... I understand all the basics of rational functions but I find it harder when adding this stuff up sometimes. If I had gotten:

    2x^2 - x^2 square root (x+1) - 6 + x^2 square root (x+1)

    how would I add it then? I made those numbers up.... but how would i add x^2 quare root (x+1) with x^2 square root (x+1) ... would it just become x^4 square root (x+1) ?


    and if i had x^2 square root (x+1) + x^3 square root (x+1) ... would i still add the two to become x^5 (square root x+1) or do i leave it since hte exponenets are different?

    thanks.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    It's okay, and it's good that you want to go through them again!

    2x^2 - x^2 square root (x+1) - 6 + x^2 square root (x+1)

    how would I add it then? I made those numbers up.... but how would i add x^2 quare root (x+1) with x^2 square root (x+1) ... would it just become x^4 square root (x+1) ?
    One way for you to be sure, you can introduce something else.

    Let y = x^2\sqrt{x+1}

    Then

    2x^2 - x^2\sqrt{x+1} - 6 + x^2\sqrt{x+1} = 2x^2 - y - 6 + y

    How would you deal with this? It's simple like that, right?

    and if i had x^2 square root (x+1) + x^3 square root (x+1) ... would i still add the two to become x^5 (square root x+1) or do i leave it since hte exponenets are different?
    x^2 \sqrt{x+1} + x^3\sqrt{x+1}

    No no. I see that you are confusing something.

    x^2 + x^3 it cannot be further simplified (other than by factoring giving x^2(1+ x)

    x^2 \times x^3 = x^5 When you multiply something with the same base (x) but different powers, then you add. If it's division, you subtract.

    Here what you can do is only factor x^2 \sqrt{x+1} and simplify:

    x^2 \sqrt{x+1} + x^3\sqrt{x+1} = x^2\sqrt{x+1}(1 + x) = x^2(x+1)^{3/2}

    Ok, I might not reply for a while now; I'm going to bed. See you tomorrow!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2009
    Posts
    123
    Thanks for the help, I appreciate it. Can you let me know if I added and subtracted the functions right or not in the original question? If i have then i'll try dividing the 2 myself before I ask any further questions.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    The reason I didn't tell you about them is that because they're correct
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. two easy questions
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: August 29th 2011, 05:12 PM
  2. some Easy Questions
    Posted in the Algebra Forum
    Replies: 44
    Last Post: November 23rd 2008, 04:15 AM
  3. Taylor series and the reminder
    Posted in the Calculus Forum
    Replies: 9
    Last Post: July 21st 2008, 10:59 AM
  4. integration reminder
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 4th 2007, 01:42 AM
  5. Easy questions that still confuse me.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 19th 2007, 01:38 PM

Search Tags


/mathhelpforum @mathhelpforum