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Math Help - functions differenciate

  1. #1
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    functions differenciate

    im terrible @ math. please help me with this!

    given 2 functions: f(x) = 2x, g(x) = -2x+5x

    im supposed to determine the following

    g(f)
    inverse of g
    g (g inverse)
    the difference quotient for f
    the average rate of change for f

    yeah... a lot of Q's. someone pls explain the process to me, cuz i fail to understand it...
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  2. #2
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    Quote Originally Posted by brokedude View Post
    im terrible @ math. please help me with this!

    given 2 functions: f(x) = 2x, g(x) = -2x+5x

    im supposed to determine the following

    g(f)
    inverse of g
    g (g inverse)
    First part:

    [g(f)](x)=g(f(x))=-2(f(x))^2+5f(x)=2(2x)^2+5(2x)=8x^2+10x<br />

    Second part:

    There is no function g^{-1}, as for most y there are two xs such that g(x)=y.

    RonL
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    Quote Originally Posted by brokedude View Post
    im terrible @ math. please help me with this!

    given 2 functions: f(x) = 2x, g(x) = -2x+5x

    im supposed to determine the following:

    the difference quotient for f
    the average rate of change for f
    The difference quotient for f is:

    <br />
DQ(f,h)=\frac{f(x+h)-f(x)}{h}=\frac{2(x+h)-2x}{h}=\frac{2h}{h}=2<br />

    Thus we see that the average rate of change of f over
    any interval (x, x+h) is 2

    RonL
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by brokedude View Post
    g(x) = -2x+5x

    im supposed to determine the following

    g(f)
    inverse of g
    g (g inverse)
    the difference quotient for f
    the average rate of change for f
    Quote Originally Posted by CaptainBlack View Post
    First part:

    Second part:

    There is no function g^{-1}, as for most y there are two xs such that g(x)=y.

    RonL
    CaptainBlack is absolutely correct about the inverse of g. It has no inverse. However we may informally get an inverse (that is to say, the process is correct even if the application is not) by:
    g(x) = -2x^2 + 5x

    Let y = -2x^2 + 5x

    Now switch the roles of x and y:
    x = -2y^2 + 5y

    Now solve for y:
    2y^2 - 5y + x = 0

    y = \frac{5 \pm \sqrt{25 - 8x}}{4} <-- via the quadratic formula

    Thus
    g^{-1}(x) = \frac{5 \pm \sqrt{25 - 8x}}{4}

    The problem, as CaptainBlack mentioned, is that for the graph y = g(x) there are two y values for every x (except at the vertex point), so we need to be very careful about defining a domain on which an inverse exists and just what that inverse is. (It will either be the "+" or the "-" of the inverse formula given above.)

    -Dan
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    thanks for your help !!
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    Quote Originally Posted by topsquark View Post
    CaptainBlack is absolutely correct about the inverse of g. It has no inverse. However we may informally get an inverse (that is to say, the process is correct even if the application is not) by:
    g(x) = -2x^2 + 5x

    Let y = -2x^2 + 5x

    Now switch the roles of x and y:
    x = -2y^2 + 5y

    Now solve for y:
    2y^2 - 5y + x = 0

    y = \frac{5 \pm \sqrt{25 - 8x}}{4} <-- via the quadratic formula

    Thus
    g^{-1}(x) = \frac{5 \pm \sqrt{25 - 8x}}{4}

    The problem, as CaptainBlack mentioned, is that for the graph y = g(x) there are two y values for every x (except at the vertex point), so we need to be very careful about defining a domain on which an inverse exists and just what that inverse is. (It will either be the "+" or the "-" of the inverse formula given above.)

    -Dan
    There is a way around the problem which is to extend g from \bold{R} to \mathcal{P}(\bold{R}), when:

    g(g^{-1})=I the identity function I(S)=S for all subsets of \bold{R}

    but I doubt that this is what is wanted.

    RonL
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