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Thread: Composite Functions

  1. #1
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    Composite Functions

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

    Pleack check and correct the following composite functions

    1) f(g(x)) = $\displaystyle \sqrt {3+3x+3x^2} $ Domain: all real numbers. Is this correct?
    2) g(f(x)) = 1+$\displaystyle \sqrt {3x}$+3x Domain: x greater than 0
    3) f(f(x)) = f($\displaystyle \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) =$\displaystyle \sqrt {3x} $
    g(x) = 1+x+x^2

    Pleack check and correct the following composite functions

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

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

    3) f(f(x)) = f($\displaystyle \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 $\displaystyle h(x)$ and $\displaystyle k(x)$

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

    an explicit example would be this:

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

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

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

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

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

    3.f(f(x)) = f($\displaystyle \sqrt {x-3}$ = ((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 $\displaystyle a^{b}$

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

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

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

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

    Here is a good post explaining fractional exponents http://www.mathhelpforum.com/math-help/51855-post3.html (it's even illustrated!)
    -----
    So your example:
    $\displaystyle (-3)^{1/4} = \sqrt[4]{-3}$, 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) = $\displaystyle \sqrt {x-3} $
    g(x) = 3/x
    1. f(g(x)) = f(3/x) = $\displaystyle \sqrt {(3-3x)/x}$
    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

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

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

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

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

    3.f(f(x)) = f($\displaystyle \sqrt {x-3}$ = ((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 $\displaystyle 0 < x \le 1$ or $\displaystyle x \in (0,1]$ or $\displaystyle \mbox{dom}(f) = (0,1]$ 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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