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Math Help - Composite Functions

  1. #1
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    Composite Functions

    f(x) =  \sqrt {3x}
    g(x) = 1+x+x^2

    Pleack check and correct the following composite functions

    1) f(g(x)) =  \sqrt {3+3x+3x^2} Domain: all real numbers. Is this correct?
    2) g(f(x)) = 1+ \sqrt {3x}+3x Domain: x greater than 0
    3) f(f(x)) = f( \sqrt {3x} ) =3^1/2 times (3x)^1/4 I do not know how to type square root of square root. Anyway, is the answer above correct? Domain: all real numbers
    4) g(g(x)) = g(1+x+x^2) = 1+(1+x+x^2)+(1+x+x^2)^2 = x^4+2x^3+4x^2+3x+3 Domain: all real numbers.

    Thank you very much
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oceanmd View Post
    f(x) =  \sqrt {3x}
    g(x) = 1+x+x^2

    Pleack check and correct the following composite functions

    1) f(g(x)) =  \sqrt {3+3x+3x^2} Domain: all real numbers. Is this correct?
    yes

    2) g(f(x)) = 1+ \sqrt {3x}+3x Domain: x greater than 0
    no, what if x = 0? would that be a problem?

    3) f(f(x)) = f( \sqrt {3x} ) =3^1/2 times (3x)^1/4 I do not know how to type square root of square root. Anyway, is the answer above correct? Domain: all real numbers
    no, what if x is negative?

    4) g(g(x)) = g(1+x+x^2) = 1+(1+x+x^2)+(1+x+x^2)^2 = x^4+2x^3+4x^2+3x+3 Domain: all real numbers.
    yes
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  3. #3
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    Jhevon,

    First of all thank you for your help.

    I still do not understand why x cannot be negative in f() =3^1/2 times (3x)^1/4. Let's say x=-1, then (3 times -1)^1/4 is 1/(-3)^4 =1/81, which will make it a real solution. Where am I going wrong?

    Thank you
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oceanmd View Post
    Jhevon,

    First of all thank you for your help.

    I still do not understand why x cannot be negative in f() =3^1/2 times (3x)^1/4. Let's say x=-1, then (3 times -1)^1/4 is 1/(-3)^4 =1/81, which will make it a real solution. Where am I going wrong?

    Thank you
    one thing you must get about composite functions is that they require the domain to work for the function being inputed.

    say we have two functions h(x) and k(x)

    then h(k(x)) means we take the output for the function k and plug it into h. but that means the x's that we picked must be those that work in k, otherwise it is meaningless.

    an explicit example would be this:

    f(x) = x^2 and g(x) = \sqrt{x}

    then f(g(x)) = (\sqrt{x})^2 = x

    what is the domain here? y = x is a function defined for all x, does that mean the domain here is all real x? No! it does not. the domain here is x \in [0, \infty), because we have to make sure the domain works for g, since the values we can plug into f comes from the output of g, and the outputs of g do not work for all x
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  5. #5
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    Jhevon,

    Got it. Will you please check one more.
    f(x) = \sqrt {x-3}
    g(x) = 3/x
    1. f(g(x)) = f(3/x) =  \sqrt {(3-3x)/x}
    Question: how to solve (3-3x)/x greater or equal 0?
    Domain: x cannot be 0 and ?

    2. g(f(x)) = g( \sqrt {x-3} = 3/ \sqrt{x-3}
    domain: x greater than 3

    3.f(f(x)) = f( \sqrt {x-3} = ((x-3)^1/2-3)^1/2
    Domain: x greater or equal 12

    4. g(g(x)) =g(3/x) = x
    Domain: x canot be 0

    Thank you
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  6. #6
    Super Member angel.white's Avatar
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    Quote Originally Posted by oceanmd View Post
    I still do not understand why x cannot be negative in f() =3^1/2 times (3x)^1/4. Let's say x=-1, then (3 times -1)^1/4 is 1/(-3)^4 =1/81, which will make it a real solution. Where am I going wrong?
    This is a bit difficult to follow, try using the latex notation for a power, which is a^{b} which returns a^{b}

    Anyway, the statement " (3 *-1)^{1/4} \mbox{ is } \frac{1}{(-3)^{4}} =\frac{1}{81}" is very wrong

    here are some basic things for you:
    a^{b/c}=\sqrt[c]{a^{b}}=\sqrt[c]{a}^{b}

    a^{-b}=\frac{1}{a^{b}}

    And combine them:
    a^{-b/c}=\frac{1}{\sqrt[c]{a^{b}}}=\frac{1}{(\sqrt[c]{a})^{b}}

    Here is a good post explaining fractional exponents http://www.mathhelpforum.com/math-help/51855-post3.html (it's even illustrated!)
    -----
    So your example:
    (-3)^{1/4} = \sqrt[4]{-3}, and what real number can you multiply by itself 4 times to get a negative number? there is none, so if you are taking the fourth root of a real number, that number cannot be negative. (note that this is only for even roots, if it were the third root, then it would be doable, because a negative number times itself 3 times will be negative)
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oceanmd View Post
    Jhevon,

    Got it. Will you please check one more.
    f(x) = \sqrt {x-3}
    g(x) = 3/x
    1. f(g(x)) = f(3/x) =  \sqrt {(3-3x)/x}
    Question: how to solve (3-3x)/x greater or equal 0?
    Domain: x cannot be 0 and ?
    recall that a fraction is positive if the numerator and denominator are the same sign. so we need to find for what x values do one or both of the following inequalities work for

    3 - 3x \ge 0 AND x > 0 (since we can't have x = 0)

    and/or 3 - 3x \le 0 AND x < 0

    find the solution that works for either or both of these systems

    2. g(f(x)) = g( \sqrt {x-3} = 3/ \sqrt{x-3}
    domain: x greater than 3

    3.f(f(x)) = f( \sqrt {x-3} = ((x-3)^1/2-3)^1/2
    Domain: x greater or equal 12

    4. g(g(x)) =g(3/x) = x
    Domain: x canot be 0

    Thank you
    the rest seems fine
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  8. #8
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    Composite functions

    Jhevon,

    Domain for f(g(x)) = f(3/x) =

    x cannot be 0, and x is greater than 0 and less or equal 1

    Is this correct?
    Thank you.
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by oceanmd View Post
    Jhevon,

    Domain for f(g(x)) = f(3/x) =

    x cannot be 0, and x is greater than 0 and less or equal 1

    Is this correct?
    Thank you.
    yes, but it is sufficient to say 0 < x \le 1 or x \in (0,1] or \mbox{dom}(f) = (0,1] or something like that. the point is, saying x is greater than zero includes x not equal zero
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  10. #10
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    Composite Functions

    Thank you, Jhevon
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