Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - Bonus Question

  1. #1
    Banned
    Joined
    Aug 2010
    From
    Singapore
    Posts
    93

    Bonus Question

    Dear friends,

    The function g is defined by g(x)= \frac {1}{(x-5)(x+3)}, x not equal to 3 and 5.

    i) Find the coordinates of the turning point of the curve y=g(x).

    ii) Hence, sketch the graph of y=g(x) and write down the range of the function g.
    Last edited by stupidguy; September 11th 2010 at 10:03 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,674
    Thanks
    1498
    Turning points occur where the derivative is 0.

    So you need to differentiate the function, set it equal to 0 and solve for x. Then back-substitute to find the corresponding y.


    To differentiate this function, you need to rewrite it as

    g = [(x-5)(x+3)]^{-1}.


    You will now need to use a combination of the Chain Rule and the Product Rule (or expand the inside and just use the Chain Rule).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Aug 2010
    From
    Singapore
    Posts
    93
    so dy/dx=\frac {2-2x}{(x^2-2x-15)^2}

    2-2x=0
    2=2x
    x=1
    y=\frac{1}{-16}

    Using Graphmatica, the answer dun match the graph.

    Would appreciate if u provide me with the full solution, then I slowly ponder on it.
    Last edited by stupidguy; August 25th 2010 at 08:01 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,674
    Thanks
    1498
    I get the same answer.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Aug 2010
    From
    Singapore
    Posts
    93
    so we get the turning point which is (1, 1/-16)......how to continue?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by stupidguy View Post
    Dear friends,

    I am too poor to afford maths tuition. Please teach me how to do this simple sum.

    The function g is defined by g(x)= \frac {1}{(x-5)(x+3)}, x not equal to 3 and 5.

    i) Find the coordinates of the turning point of the curve y=g(x).

    ii) Hence, sketch the graph of y=g(x) and write down the range of the function g.
    *Ahem* Attention all posters (including the original poster):

    Since this question is posted in the PRE-calculus subforum, the expectation is that the question is to be done without using calculus.

    Draw the graph of the quadratic function y = (x - 5)(x + 3). Now think about what happens to this graph when you take the reciprocal of the y-coordinates for each x-coordinate.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,674
    Thanks
    1498
    Quote Originally Posted by mr fantastic View Post
    *Ahem* Attention all posters (including the original poster):

    Since this question is posted in the PRE-calculus subforum, the expectation is that the question is to be done without using calculus.

    Draw the graph of the quadratic function y = (x - 5)(x + 3). Now think about what happens to this graph when you take the reciprocal of the y-coordinates for each x-coordinate.
    And yet, when given the advice to use the derivative, the OP knew exactly what to do...

    This is why I say there should be a link to the Calculus forum in the High School section of this forum...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Aug 2010
    From
    Singapore
    Posts
    93
    @Prove It: good suggestion.

    @ mrfantastic sry, I do not know the difference between calculus and pre-calculus. However, I I just want someone to teach me how to tackle this question. I dun get it ur "advice"
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,674
    Thanks
    1498
    Think about \frac{1}{n}.

    What happens to \frac{1}{n} when n is small? What happens to \frac{1}{n} as n gets large?


    In this case, n = (x - 5)(x + 3).
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by stupidguy View Post
    @Prove It: good suggestion.

    @ mrfantastic sry, I do not know the difference between calculus and pre-calculus. However, I I just want someone to teach me how to tackle this question. I dun get it ur "advice"
    (x-5)(x+3)

    has zeros at x=5

    and x=-3

    It's a U-shaped graph and the minimum occurs halfway between the zeros.
    If you invert the graph to

    \displaystyle\frac{1}{(x-5)(x+3)}

    we now have vertical asymptotes at x=5 and x=-3

    but the turning point still occurs at the same x we previously had the minimum.
    This is because..

    \displaystyle\frac{1}{most\ negative}=least\ negative

    or

    \displaystyle\frac{1}{least\ positive}=most\ positive
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Banned
    Joined
    Aug 2010
    From
    Singapore
    Posts
    93
    If you invert the graph to



    we now have vertical asymptotes at and

    but the turning point still occurs at the same x we previously had the minimum.
    This is because..



    or
    still dun get the link. have not clearly answered my initial question.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Banned
    Joined
    Aug 2010
    From
    Singapore
    Posts
    93

    graph

    Bonus Question-untitled.jpg

    attached is the graph....but i need to know how to derive it.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by stupidguy View Post
    Click image for larger version. 

Name:	Untitled.jpg 
Views:	6 
Size:	286.5 KB 
ID:	18723

    attached is the graph....but i need to know how to derive it.
    The black parabola is the graph of

    (x-5)(x+3)

    The pink asymptotic curve is the graph of

    \displaystyle\frac{1}{(x-5)(x+3)}

    The y-axis is not suitably scaled to effectively view the pink curve.
    If we "stretch" it, you'd more easily see that the local maximum (turning point) of the pink graph
    occurs at the exact same x which locates the minimum value (turning point) of the U-shaped parabola.

    If you invert a quadratic equation, the zeros go towards infinity (become vertical asymptotes),
    local maxima become local minima and local minima become local maxima.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Banned
    Joined
    Aug 2010
    From
    Singapore
    Posts
    93
    not clear. I can only understand that x=5 and x=-3 are vertical asymptotes, (1,-16) is the max point without the use of the U-shaped parabola.

    Someone end this misery by giving me the complete steps and answers to my question. Dun just hint!
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,674
    Thanks
    1498
    Quote Originally Posted by stupidguy View Post
    not clear. I can only understand that x=5 and x=-3 are vertical asymptotes, (1,-16) is the max point without the use of the U-shaped parabola.

    Someone end this misery by giving me the complete steps and answers to my question. Dun just hint!
    We are not a homework service. You need to make an effort to understand this material yourself.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Bonus question involving derivatives?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 20th 2010, 06:14 PM
  2. Very Hard Bonus Question
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: February 24th 2010, 10:24 PM
  3. Bonus Question involving Continuity?
    Posted in the Calculus Forum
    Replies: 17
    Last Post: February 18th 2010, 09:52 AM
  4. Calculus Class Bonus Question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 1st 2009, 02:26 PM
  5. Bonus Logarithem Question
    Posted in the Algebra Forum
    Replies: 3
    Last Post: December 7th 2006, 06:58 PM

Search Tags


/mathhelpforum @mathhelpforum