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Thread: a lot of confusion about Functions!!

  1. #1
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    a lot of confusion about Functions!!

    consider the following functions

    a lot of confusion about Functions!!-maths.png

    I have attached a screen shot of my problem as that was easier than typing it out.

    Apologies if this is posted in the wrong section, i have to do this problem by Thursday and to be honest i don't really have a clue but its important i get this right. anyway ill take a stab at it and hopefully some more experienced people can set me right.

    so for the first part with the graph i need to get f(x) = g(x)
    which would look something like this


    f(x)=x2-1
    g(x)=x+1

    f(x) = g(x)
    x2-1=x+1
    (x+1)-(x2-1)=o
    (x+1)-x2+1=0
    -x2+x+2=0


    S0 from here i need to get some co-ordinates from the above equation so i can use to draw a line on the graph?? i think, im not really sure where to go from here....and my maths may even be way off here like i said im unsure

    moving on to next section
    compute the following values:
    f(2)
    g(0)
    f(2/3)


    f(2)=(2)2-1 = 4-1=3
    g(0)=(0) +1=1
    f(2 over 3)= (2 over 3)2+1= ??

    i think the first two may be right not sure what to do with fraction in last one.

    as for the last section

    f dot g (x) = f(x+1)=(x+1)2-1
    g dot f (x) = g(x2-1) = (x2-1)+1
    f dot (x) = f(x2-1) = (x2-1)2-1
    g dot g(x) = g(x+1)=(x+1) + 1

    again i think im right here? could some one confirm if i am or not

    I realize this is a long post i just want to thank anyone in advance who can lend me their expertise!
    Last edited by ronanbrowne88; Oct 22nd 2013 at 10:38 AM.
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  2. #2
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    Re: a lot of confusion about Functions!!

    Quote Originally Posted by ronanbrowne88 View Post
    so for the first part with the graph i need to get f(x) = g(x)
    which would look something like this

    f(x)=x2-1
    g(x)=x+1

    f(x) = g(x)
    x2-1=x+1
    (x+1)-(x2-1)=o
    (x+1)-x2+1=0
    -x2+x+2=0
    This is correct so far. I would recommend multiplying both sides of the equation by -1 and then trying to factor the polynomial.

    Quote Originally Posted by ronanbrowne88 View Post
    S0 from here i need to get some co-ordinates from the above equation so i can use to draw a line on the graph?? i think, im not really sure where to go from here....and my maths may even be way off here like i said im unsure
    That is correct. Making a table of values might help.

    Quote Originally Posted by ronanbrowne88 View Post
    moving on to next section
    compute the following values:
    f(2)
    g(0)
    f(2/3)

    f(2)=(2)2-1 = 4-1=3
    g(0)=(0) +1=1
    f(2 over 3)= (2 over 3)2+1= ??
    You are correct for the first two. For the last one, an exponent of 2 means multiply the number by itself. Do you remember how to multiply two fractions? I will set it up for you:

    $\displaystyle f\left( \dfrac{2}{3}\right) = \left(\dfrac{2}{3}\right)^2+1 = \dfrac{2}{3}\cdot \dfrac{2}{3} + 1$

    Quote Originally Posted by ronanbrowne88 View Post
    as for the last section ill i don't know how to even attempt it if anyone could show me how to do one as a example i can try the rest.

    I realize this is a long post i just want to thank anyone in advance who can lend me their expertise!
    The last section is composition of functions. So, $\displaystyle (f\circ g)(x)$ means take the function $\displaystyle f(x)$ and wherever you see an $\displaystyle x$, replace it with $\displaystyle g(x)$. So, $\displaystyle (f\circ g)(x) = f(g(x)) = (g(x))^2+1$. Now, in place of $\displaystyle g(x)$, put the formula for $\displaystyle g(x)$.
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  3. #3
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    Re: a lot of confusion about Functions!!

    thanks for the advice, so continuing on with your advice in mind first section

    f(x)=x2-1
    g(x)=x+1

    f(x) = g(x)
    x2-1=x+1
    (x+1)-(x2-1)=o
    (x+1)-x2+1=0
    -x2+x+2=0
    -x2+x+2=0 (next multiply by -1)

    x2 -x -2 =0 (next factorize)

    x2−x−2=1(x−(−1))(x−2)
    x2−x−2=(x+1)(x−2)
    x1=−1
    x2=2

    then i can use these point to draw the graph. correct??

    the last one on part 2

    f(2 over 3)= (2 over 3)2+1= 4 over 9 +1 = 1 and 4 over 9 (sorry cant see a button to write in correct format)

    and last bit

    f dot g (x) = f(x+1)=(x+1)2-1
    g dot f (x) = g(x2-1) = (x2-1)+1
    f dot (x) = f(x2-1) = (x2-1)2-1
    g dot g(x) = g(x+1)=(x+1) + 1
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  4. #4
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    Re: a lot of confusion about Functions!!

    Quote Originally Posted by ronanbrowne88 View Post
    f(x)=x2-1
    g(x)=x+1

    f(x) = g(x)
    x2-1=x+1
    (x+1)-(x2-1)=o
    (x+1)-x2+1=0
    -x2+x+2=0
    -x2+x+2=0 (next multiply by -1)

    x2 -x -2 =0 (next factorize)

    x2−x−2=1(x−(−1))(x−2)
    x2−x−2=(x+1)(x−2)
    x1=−1
    x2=2

    then i can use these point to draw the graph. correct??
    Those points only tell you the x-coordinate where the two functions intersect. To graph the functions, you will want a table with more values for each function.


    Quote Originally Posted by ronanbrowne88 View Post
    the last one on part 2

    f(2 over 3)= (2 over 3)2+1= 4 over 9 +1 = 1 and 4 over 9 (sorry cant see a button to write in correct format)
    Seems we wrote down the wrong formula for $\displaystyle f(x)$. At the top, you have $\displaystyle f(x) = x^2-1$, so it should be $\displaystyle \dfrac{4}{9} - 1$. Find a common denominator, and you can simplify.

    Quote Originally Posted by ronanbrowne88 View Post
    and last bit

    f dot g (x) = f(x+1)=(x+1)2-1
    g dot f (x) = g(x2-1) = (x2-1)+1
    Good, but you can simplify $\displaystyle (g\circ f)(x)$ (-1+1 = 0).

    Quote Originally Posted by ronanbrowne88 View Post
    f dot (x) = f(x2-1) = (x2-1)2-1
    g dot g(x) = g(x+1)=(x+1) + 1
    You can also simplify $\displaystyle (g\circ g)(x)$ (1+1 = 2).

    Other than that, everything looks great.
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  5. #5
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    Re: a lot of confusion about Functions!!

    great thanks!
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