1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fun with functions . help requested

  1. Jul 26, 2005 #1
    I have a lot of questions and I know I'm supposed to show all my work but the post would be insanely long, so I will show you what answers I have come up with on the ones that I managed to get that far. I know how to do these but I often get hung up on the fractions and trying to get rid of the exponents

    For functions:
    f(x)= (x+1)/(x-1)
    g(x)= 1/x

    Find:

    A. f+g
    I got 2x+2/x^2-x

    B. f-g
    I got 0

    C. f*g
    I got x+1/x^2-x

    D. f/g
    I got x^2+1/x-1

    E. f o g (x)
    I got 2x/1-x


    These ones I am having difficulty with, sorry for not showing my work, some answers would be very much appreciated though.

    Multiply:
    a. (x^2-2)^3

    b. (x^2+3x+5)(x+2)

    c. (2x^2-6x)(2x-4)

    Divide and define restrictions:
    a. (2x^3-x^2+2x-3) / (x-1)

    b. (x^5-x^4-4x^3+8x^2-32x+48) / (x^2+4)

    c. (x^5-1) (x-1)

    Thanks again for the help, hopefully this is just the forum I was looking for!
     
  2. jcsd
  3. Jul 27, 2005 #2

    lurflurf

    User Avatar
    Homework Helper

    For functions:
    f(x)= (x+1)/(x-1)
    g(x)= 1/x

    First thing you need to use parenthesis to make things clear. Check your steps in these, I could not as you did not post them. An easy check is to remember function equallity means equallity for all function. values.
    f(-1)=0 g(-1)=-1
    Find:

    A. f+g
    I got 2x+2/x^2-x
    no
    (f+g)(-1)=f(-1)+g(-1)=-1
    I would really like to see the steps that led to this, and where if anywhere you intended parenthesis
    B. f-g
    I got 0
    no
    (f-g)(-1)=f(-1)-g(-1)=0-(-1)=1
    C. f*g
    I got x+1/x^2-x
    yes if you meant (x+1)/(x^2-x)
    (f*g)(-1)=f(-1)*g(-1)=0*(-1)=0
    ((-1)+1)/((-1)^2-(-1))=0
    D. f/g
    I got x^2+1/x-1
    no
    (f/g)(-1)=f(-1)/g(-1)=0/(-1)=0
    E. f o g (x)
    I got 2x/1-x
    no
    (f o g)(-1)=f(g(-1))=f(-1)=0
    2(-1)/1-(-1)=-2+1=-1
    2(-1)/(1-(-1))=-1
    These ones I am having difficulty with, sorry for not showing my work, some answers would be very much appreciated though.

    Multiply:
    a. (x^2-2)^3

    b. (x^2+3x+5)(x+2)

    c. (2x^2-6x)(2x-4)
    For these remember distribution
    example
    (3x+2)(7x-5)
    =7x(3x+2)-5(3x+2)
    =21x^2+14x-15x-10
    =21x^2-x-10
    a(b+c)=a*b+a*c and commutitivity a*b=b*a
    Divide and define restrictions:
    a. (2x^3-x^2+2x-3) / (x-1)

    b. (x^5-x^4-4x^3+8x^2-32x+48) / (x^2+4)

    c. (x^5-1) (x-1)
    use polynomial long division, factor, or rearrange to get thing that will divide
    Some would restrict if the denominator is zero, likely you only want to restric if the order of the numerator is less than of the denominator. ie (x+1)/(x-1) restrict x=1 but (x-1)^4/(x-1)^3 do not. Check your book.
    Here is an example
    (x^3+3x^2+3x+1)/(x+1)
    =(x^3+x^2+2x^2+3x+1)/(x+1)
    =(x^2(x+1)+2x^2+2x+x+1)/(x+1)
    =(x^2(x+1)+2x(x+1)+(x+1))/(x+1)
    =(x^2+2x+1)(x+1)/(x+1)
    =x^2+2x+1
     
  4. Jul 27, 2005 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Show us your work at least for the first ones. As lurflurf told you, you have every one of them wrong. It should at least be obvious that since f and g are not the same, f- g cannot be 0!
     
  5. Jul 27, 2005 #4
    Thanks for the help, although I am a bit confused. These numbers you came up with are the numbers for which the domain cannot be, correct? Is that the final answer I should be coming up with?
     
  6. Jul 27, 2005 #5

    lurflurf

    User Avatar
    Homework Helper

    I sugested that when combining function, until you are more comfortable, that you check your self by using some numerical examples. Function equallity is defined by equallity for all numbers in the domain. So you can check if your answer is right. i.e. if (x+1)(x-1)=x^2-1 then this holds for all numbers. If a mistake is made and you think (x+1)(x-1)=x^2+1 it can be seen that this is an error because if x=2 (2+1)(2-1)=2 but 2^2+1=5. Also when combining functions make sure to exclude any problems from the domain. In problems like these the main thing to look for is zeros in denominators. See what your book says about cancelation, in many books something like x/x=1 is considered ok for all x since the numerator and denominator are zero together when x=0. You should look for mistakes in your work, and post some of you can not find them. So to combine functions just write the functions out with the desires operation, and simplify as needed. When multiplying remember distributive rules. When dividing flip and multiply. When dividing think about when denominators are zero.
    Final answer should be the combined function and any domain restrictions that are appropriate.
    ie
    f(x)=(x+2) g(x)=(x-3)
    (f/g)(x)=f(x)/g(x)
    (f/g)(x)=(x+2)/(x-3)
    so domain is all real numbers except 3
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fun with functions . help requested
  1. Requesting Help (Replies: 2)

Loading...