Results 1 to 12 of 12
Like Tree3Thanks
  • 1 Post By guinster
  • 1 Post By guinster
  • 1 Post By guinster

Math Help - Calculus Graph Sketching strategy.

  1. #1
    Junior Member
    Joined
    May 2014
    From
    UK
    Posts
    38

    Calculus Graph Sketching strategy.

    Hello all,

    I'm new here and I taking a course in calculus what is totally new to me, I having been work trough differentiation and now we got to do some graph sketching following 6 steps described in the book I have started this one but I'm not sure if I'm doing it correctly so all explanation and support in how to tackle and understand it is very much appreciated. So here we go

    The Function f(x)= 6x-5 / 49x2

    STEP 1 - Identify the domain of f

    I came up with the domain is all R once 49x2 is never 0.

    Is it correct?

    STEP 2 - Wether f is even or odd

    following my examples:

    f(-x) = 6(-x)-5 / 4-9(-x)2
    = -6x-5 / 4-9(x)2

    Thus -f(x) the function is odd

    STEP 3 - x and y intercepts

    The x intercept is 6x-5=0
    6x=5
    x=5/6
    x = 0.833
    The y intercept is f(0) = 6(0)-5 / 4-9(0)2
    = -5 / 4
    = -1.25

    STEP 4 - The Intervals where x is positive or negative

    f(x) = 6x-5 / 4-9x2
    = 6x-5 / (2-3x)(2+3x)

    x (-∞,-1.25) -1.25 (-1.25,0.833) 0.833 (0.833,∞)
    6x-5 - - - - +
    2-3x + - - - -
    2+3x - + + + +
    f(x) + + + + -

    Thus f is positive on the Interval (-∞,-1.25) and negative on the interval(0.833,∞).

    STEP 5
    - The Intervals where f is increasing or decreasing and any stationary points, local maxima and local minima.

    f ' (x) = (6)(-9x2+4) - (-18x)(6x-5) / (4-9x2)2

    = 54x-66 / (4-9x2)2

    x (-∞,-1.25) -1.25 (-1.25,0.833) 0.833 (0.833,∞)
    54x2-66 - + - - +
    (4-9x2)2 + + + + +
    f ' (x) - + - - +

    Thus f is decreasing on
    (-∞,-1.25)
    and increasing on
    (0.833,∞)
    also f has a stationary point at
    0.833
    which is a local minim with value f(0.833) = 0 and a stationary point at -1.25 which is a local maximum with value f(-1.25) = 1.242

    STEP 6 - The asymptotic behaviour of f.

    I can't understand how to discover the asymptote behaviours.


    Well this is it does it all make sense to anyone? If it does what I have done is correct? If not please I really need to understand it and a step by step will really be appreciated.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    May 2014
    From
    UK
    Posts
    38

    Re: Calculus Graph Sketching strategy.

    Please know one can give me some advices so I know if it is wrong or right?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,956
    Thanks
    1780
    Awards
    1

    Re: Calculus Graph Sketching strategy.

    Quote Originally Posted by michele View Post
    Please know one can give me some advices so I know if it is wrong or right?
    Frankly. I for one find your notation so difficult to read. So I just did not.

    In fact, I not even sure what the function is. Is it $f(x)=\dfrac{6x-5}{4-9x^2}~?$

    Clearly that in a lot easier to read. Learning to use LaTeX.

    The code \$f(x)=\dfrac{6x-5}{4-9x^2}\$ gives $f(x)=\dfrac{6x-5}{4-9x^2}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2014
    From
    Britain
    Posts
    18
    Thanks
    4

    Re: Calculus Graph Sketching strategy.

    Hi
    I think you may have missed the asymptotes on the y axis where f is undefined.
    These occur where the denominator tends to zero.
    If you solve 4-9x^2 = 0 it gives +/- 2/3.
    So domain is R - {-2/3 , 2/3} and graph approaches infinity either side of these two points like 1/x does at x = 0.
    Thanks from michele
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jun 2014
    From
    Britain
    Posts
    18
    Thanks
    4

    Re: Calculus Graph Sketching strategy.

    Hi
    If you draw two vertical lines on your graph - one at y = -2/3 and the other at y = 2/3 these are the asymptotes.
    The curve will approach either infinity or - infinity either side of the asymptote.

    I have done some sketches.

    Hope they help.

    Calculus Graph Sketching strategy.-asym-2.jpg

    Regards
    Thanks from michele
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    May 2014
    From
    UK
    Posts
    38

    Re: Calculus Graph Sketching strategy.

    My sketch looks similar to yours

    But when I checked on my pc the graph should look like this:
    Is it correct?
    Attached Thumbnails Attached Thumbnails Calculus Graph Sketching strategy.-screen-shot-2014-06-26-00.21.16.jpg  
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1

    Re: Calculus Graph Sketching strategy.

    I'm coming in late to the party. Step 2... \frac{6x - 5}{4 - 9x^2} is neither even nor odd, as your graph shows.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    May 2014
    From
    UK
    Posts
    38

    Re: Calculus Graph Sketching strategy.

    An how can I prove it is neither even or odd?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1

    Re: Calculus Graph Sketching strategy.

    The same way you did. You just didn't come to the right conclusion.

    A function is even if f(-x) = f(x). A function is odd if f(-x) = -f(x). In your case:
    f(x) = \frac{6x - 5}{4 - 9 x^2} implies that

    f(-x) = \frac{6(-x) - 5}{4 - 9 (-x)^2}

    f(-x) = \frac{-6x - 5}{4 - 9 x^2} \neq f(x) \text{ nor } -f(x)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    May 2014
    From
    UK
    Posts
    38

    Re: Calculus Graph Sketching strategy.

    Thanks topsquark, so with this conclusion my graph now is valid?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Jun 2014
    From
    Britain
    Posts
    18
    Thanks
    4

    Re: Calculus Graph Sketching strategy.

    Hi

    My sketches (A) to (D) where meant to cover the four possible types of asymptote.

    Your graph is showing that the behaviour at x = -2/3 is as in my sketch (D) and
    the behaviour at x = 2/3 is as in my sketch (C).

    Regards
    Thanks from michele
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    May 2014
    From
    UK
    Posts
    38

    Re: Calculus Graph Sketching strategy.

    Thanks Again Guisnster.

    Now I fell more confident about my work.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Using the graph sketching strategy
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 2nd 2014, 08:25 PM
  2. Replies: 3
    Last Post: November 17th 2012, 06:41 PM
  3. calculus homework help... sketching a graph of f
    Posted in the New Users Forum
    Replies: 1
    Last Post: November 12th 2012, 07:08 PM
  4. Sketching A Graph
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: October 9th 2011, 11:50 AM
  5. Graph sketching
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: June 3rd 2007, 06:38 AM

Search Tags


/mathhelpforum @mathhelpforum