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Math Help - functions

  1. #1
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    functions

    am having some problems with these equations

    given tha f and g are functions where the target set and source sets are the set of real numbers and

    f(x) = 10-3x
    g(x) = 2
    1+3

    calculate

    i) f(2)
    ii)g(-3)
    iii) fog(2)

    write down an expression giving
    iv) g o f (x)
    v)f-1(x)

    again thanks for any help
    Last edited by red55; May 20th 2007 at 06:40 AM.
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  2. #2
    Member Glaysher's Avatar
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    Quote Originally Posted by red55 View Post
    am having some problems with these equations

    given tha f and g are functions where the target set and source sets are the set of real numbers and

    f(x) = 10-3x
    g(x) = 2/1+3
    Sure you've typed g correctly?
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  3. #3
    Junior Member ginafara's Avatar
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    Quote Originally Posted by red55 View Post
    am having some problems with these equations

    given tha f and g are functions where the target set and source sets are the set of real numbers and

    f(x) = 10-3x
    g(x) = 2
    1+3

    calculate

    i) f(2)
    ii)g(-3)
    iii) fog(2)

    write down an expression giving
    iv) g o f (x)
    v)f-1(x)

    again thanks for any help
    i) f(2) where 2 is your x value... this is the function f(x) evaluated at 2 so you plug in 2 where ever you have an x in your function

    f(2) = 10-3(2) = 4

    ii) Unforutnately your g function is not reasonable, I think there is an error...

    but g(-3) would be done the same way, substitute x with -3 because you are evaluating the function g(x) at -3

    iii) the composite is not as scary as it seems...(well depending on your functions anyway...)
    f(g(x)) means that the functions f(x) is evaluated at g(x) so substitute x in f(x) with g(x)...

    you will have

    f(g(x)) = 10 - 3(g(x))

    iv) g compositite f is the same as iii), but instead you use f(x) as your x-value and plug into all x's in g(x)

    v) f-1 is f inverse.

    Take your function,

    f(x) = 10 - 3x Solve for x

    f(x)= 10 - 3x ---> move the 3x to the left side of the = sign and the f(x) to the right of the = sign

    10 - f(x) = 3x ----> get x by itself by dividing both sides by 3

    (10 - f(x))/3 = x -----> no that you have this just swap x with f(x)

    (10 - x)/3 = f(x) ------> this is your inverse function

    I hope that helps
    Last edited by ginafara; May 25th 2007 at 11:54 PM. Reason: Used y in part v, but thought I should be consistent with the orginal problem...
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  4. #4
    Super Member
    earboth's Avatar
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    Quote Originally Posted by red55 View Post
    am having some problems with these equations

    given tha f and g are functions where the target set and source sets are the set of real numbers and

    f(x) = 10-3x
    g(x) = 2
    1+3

    calculate

    i) f(2)
    ii)g(-3)
    iii) fog(2)

    write down an expression giving
    iv) g o f (x)
    v)f-1(x)

    again thanks for any help
    Hello,

    I assume that the function g reads:
    g(x) = \frac{2}{x+3}

    If so you'll get at:

    ii) g(-3) \notin \mathbb{R} because you have to divide by zero.

    iii) f \circ g(2)=10 - 3 \cdot \left( \frac{2}{2+3} \right) = 10 -\frac{6}{5} = \frac{44}{5}

    iv) g \circ f(x) = g(f(x))=\frac{2}{10 - 3x +3} = \frac{2}{13-3x}
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