Results 1 to 5 of 5

Math Help - Graph of a function, revision please help

  1. #1
    Junior Member
    Joined
    May 2006
    Posts
    31

    Graph of a function, revision please help

    a) Explain why the graph of the function f(x) = (x-3)(e^-x/4 - 1) lies above the x- axis for 0<x<3 and below x-axis for x>3

    b) Use this fact to find the area enclosed by the graph and the x-axis between x = 0 and x = 6, giving your answer to 4 d.p
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by fair_lady0072002
    a) Explain why the graph of the function f(x) = (x-3)(e^-x/4 - 1) lies above the x- axis for 0<x<3 and below x-axis for x>3
    You have,
    f(x)=\frac{x-3}{e^{-x/4}-1}
    You want to find where it is below the x-axis thus, you want to find,
    f(x)<0
    Thus,
    \frac{x-3}{e^{-x/4}-1}<0
    You are dividing two fractions.
    They are negative when they have opposite signs,
    x-3<0 \mbox{ and }e^{-x/4}-1 >0 (1)
    Or,
    x-3>0 \mbox{ and }e^{-x/4}-1 <0 (2)
    ---
    Now solve both inequalities.
    (1) The first part is easy x-3<0 thus, x<3
    Now, the second part,
    e^{-x/4}-1>0
    Thus,
    e^{-x/4}>1
    Okay, you need to know something about exponential curves.
    An exponential curve a^x is increasing and intersectest y-axis at x=0 An exponential curve a^{-x} is decreasing and intersectest y-axis at x=0.

    This exponent e^{-x/4} is of the type a^{-x} So it is decreasing and intersects the y-axis (which is one) at x=0 Thus, when x>0 The function (since decreasing) is always less than one.

    You now have for the first part x<3 \mbox{ and }x>0, a more fancy way to say this is 0<x<3.
    ---
    For the second part (2), following same ideas you arrive at,
    x>3 \mbox{ and } x<0, but this set is empty (basically impossible, right?) Thus, you disregard this case.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by fair_lady0072002
    b) Use this fact to find the area enclosed by the graph and the x-axis between x = 0 and x = 6, giving your answer to 4 d.p
    Whoa!
    Remember on the interval [0,6] the function is not countinous (discountinous at x=0). Since that point is a vertical asymptote, This is a Type II Improper Integral.
    Furthermore, It is negative on (0,3) and positive on (3,6]
    Therefore you need to do the integrals
    \int_{0^+}^3 \frac{x-3}{e^{-x/4}-1} dx-\int_3^6 \frac{x-3}{e^{-x/4}-1}dx
    The reason why a negative because the function is negative there.
    And also remember that,
    \int_{0^+}_3 \frac{x-3}{e^{-x/4}-1} dx=\lim_{N\to 0^+}\int_{N}^3 \frac{x-3}{e^{-x/4}-1} dx
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Quote Originally Posted by fair_lady0072002
    a) Explain why the graph of the function f(x) = (x-3)(e^-x/4 - 1) lies above the x- axis for 0<x<3 and below x-axis for x>3

    b) Use this fact to find the area enclosed by the graph and the x-axis between x = 0 and x = 6, giving your answer to 4 d.p
    Isn't that multiplication instead of division?.
    Last edited by galactus; November 24th 2008 at 06:39 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by galactus
    Isn't that multiplication instead of division?.
    Why do I always do that?

    (Does that mean that this would tear a hole in the spacetime countinuum)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 26th 2009, 05:51 AM
  2. Revision
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 25th 2009, 02:04 AM
  3. revision help
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 8th 2009, 10:41 AM
  4. Some Revision Help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 27th 2008, 07:45 PM
  5. Revision
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 2nd 2007, 07:21 AM

Search Tags


/mathhelpforum @mathhelpforum