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Math Help - Verify that x = 1 => fg(x) = gf(x)

  1. #1
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    Verify that x = 1 => fg(x) = gf(x)

    a)The functions f and g are defined by f(x) = root x for x > 0; g(x) = x - 1 for all values of x.

    i.)Write down expressions for fg(x) and gf(x).
    > fg(x) = root x - 1
    > gf(x) = root x - 1

    ii.) Verify that x = 1 => fg(x) = gf(x)
    >??????
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  2. #2
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    Quote Originally Posted by ansonbound View Post
    a)The functions f and g are defined by f(x) = root x for x > 0; g(x) = x - 1 for all values of x.

    i.)Write down expressions for fg(x) and gf(x).
    > fg(x) = root x - 1
    > gf(x) = root x - 1

    ii.) Verify that x = 1 => fg(x) = gf(x)
    >??????
    f(x) = \sqrt{x} , x > 0

    g(x) = x - 1

    I assume you mean function composition by the notation "fg(x) and gf(x)"

    f[g(x)] = f(x-1) = \sqrt{x-1}

    g[f(x)] = g(\sqrt{x}) = \sqrt{x} - 1

    you should be able to see that f[g(1)] = g[f(1)] = 0
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  3. #3
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    Quote Originally Posted by skeeter View Post
    f(x) = \sqrt{x} , x > 0

    g(x) = x - 1

    I assume you mean function composition by the notation "fg(x) and gf(x)"

    f[g(x)] = f(x-1) = \sqrt{x-1}

    g[f(x)] = g(\sqrt{x}) = \sqrt{x} - 1

    you should be able to see that f[g(1)] = g[f(1)] = 0
    so it is right that i say \sqrt{x-1} = \sqrt{x} - 1 = 0?
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  4. #4
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    Quote Originally Posted by ansonbound View Post
    so it is right that i say \sqrt{x-1} = \sqrt{x} - 1 = 0?
    not for all x ... only for x = 1
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    the last question..
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  6. #6
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    y = \sqrt{x-1}

    swap variables ...

    x = \sqrt{y-1}

    solve for y to finish
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  8. #8
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    Quote Originally Posted by ansonbound View Post
    yes! i did it by using
    b^2 - 4ac = 0<br />
16 - 4*k*2 = 16<br />
k = 2
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  9. #9
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    sketch the graph y = |2x-4|

    sketch the graph y = x

    how far up will you have to translate y = |2x-4| so that it touches y = x at only one point?
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