Results 1 to 7 of 7

Math Help - Double Division?

  1. #1
    Senior Member
    Joined
    Jul 2007
    Posts
    290

    Double Division?

    Find f(x) / g(x)

    F(x) = 2 / x

    G(x) = 4 / (x + 4)

    (2/x) / (4/x+4)


    I kind of forgot how to tackle these kinds of problems


    Oops nevermind I figured it out. Reciprocal and multiply through. Ok more important question, the result of the problem will be (x + 8) / (2x) . According the back of the book, the domain is x cannot equal 0 and -4. The thing is, clearly -4 can be used in the domain seeing as it will result in -8 in the denominator which is ok. Can someone clear this up?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    \frac{\dfrac2x}{\dfrac4x+4}=\frac{x\cdot\dfrac2x}{  x\cdot\left(\dfrac4x+4\right)}=\frac2{4+4x}=\frac1  {2+2x}

    Of course, \forall x\ne0,-1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Quote Originally Posted by Krizalid View Post
    \frac{\dfrac2x}{\dfrac4x+4}=\frac{x\cdot\dfrac2x}{  x\cdot\left(\dfrac4x+4\right)}=\frac2{4+4x}=\frac1  {2+2x}

    Of course, \forall x\ne0,-1


    I got (x + 8) / (2x) for the final answer which is the same as the answer in the back of the book
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Hey, but you changed the question!

    \frac{{f(x)}}<br />
{{g(x)}} = \frac{{\dfrac{2}<br />
{x}}}<br />
{{\dfrac{4}<br />
{{x + 4}}}} = \frac{{2\left( {x + 4} \right)}}<br />
{{4x}} = \frac{{x + 4}}<br />
{{2x}}

    Is it clear now?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Quote Originally Posted by Krizalid View Post
    Hey, but you changed the question!

    \frac{{f(x)}}<br />
{{g(x)}} = \frac{{\dfrac{2}<br />
{x}}}<br />
{{\dfrac{4}<br />
{{x + 4}}}} = \frac{{2\left( {x + 4} \right)}}<br />
{{4x}} = \frac{{x + 4}}<br />
{{2x}}

    Is it clear now?
    Thanks!!! I just checked with a friend's answer and he got the same one you did. Book was wrong . Think you could tell me about why the domain for (x+4) / (2x) is X cannot = 0 and -4? If you plug in -4 to the equation, you don't get an undefined in the denominator so i'm a bit confused as to why -4. Thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    f(x)=\frac2x\implies\text{Dom}\,f=x\in\mathbb R-\{0\}

    g(x)=\frac4{x+4}\implies\text{Dom}\,g=x\in\mathbb R-\{-4\}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Jul 2007
    Posts
    290
    Quote Originally Posted by Krizalid View Post
    f(x)=\frac2x\implies\text{Dom}\,f=x\in\mathbb R-\{0\}

    g(x)=\frac4{x+4}\implies\text{Dom}\,g=x\in\mathbb R-\{-4\}

    So in composition, the domain is the original functions before they are put together?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Division?
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: November 21st 2010, 09:10 AM
  2. Simplify Equation- Double Division with Radicals
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: November 17th 2010, 08:42 AM
  3. Division
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 18th 2010, 12:50 PM
  4. Polynomial division vs synthetic division
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 9th 2009, 06:49 AM
  5. division
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 13th 2009, 09:08 AM

Search Tags


/mathhelpforum @mathhelpforum