1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sketch the graphs of the functions - Calculus question

  1. May 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Sketch the graphs of the functions. Indicate intervals on which the function is increasing, decreasing, concave up, or concave down; indicate relative maximum points, relative minimum points, points of inflection, horizontal asympototes, vertical asymptotes, symmetry, and those intercepts that can be obtained conveniently:

    2. Relevant equations
    y=x^4 - 4x^3 - 20x^2 +150


    3. The attempt at a solution
    This is the solution I have:
    This is a polynomial of degree 4, so there are no asymptotes.
    The function has no symmetries.
    dy/dx = 4x^3 - 12x^2 - 40x
    d^2y/dx^2 = 12x^2 - 24x - 40

    dy/dx = 4(x^3 - 3x^2 - 10x) = 4x(x-5)(x+2)

    There are 0s at x = -2, 0, and 5
    Thus, there are three critical points for a 4th degree polynomial.
    As the function goes to infinity as x goes to plus or minus infinity, we know that
    the function decreases on (-infinity, -2), increases on (-2, 0), decreases on (0,5), and increases on (5,infinity)
    f(-2) = (-2)^4 -4*(-2)^3 - 20*(-2)^2 + 150 = 16 + 32 - 80 + 150 = 118
    There is a local minimum at (-2, 118)
    f(0) = 150
    There is a local maximum at (0, 150)
    f(5) = (5)^4 -4*(5)^3 - 20*(5)^2 + 150 = 625 - 500 - 500 + 150 = -225
    There is a local minimum at (5, -225)

    The y-intercept is 150

    The 0s are approximately 2.5139 and 6.5252

    Finally, as the second derivative is 12x^2 - 24x - 40 = 4(3x^2 - 6x - 10), the inflection points are
    1 plus or minus sqrt(39)/3 is approximately -1.08166599946613 and 3.08166599946613

    As the second derivative is quadratic with a positive leading coefficient, we then know that
    the function is concave up on (-infinity, 1 - sqrt(39)/3) or (-infinity, -1.08166599946613 ) and
    (1 + sqrt(39)/3,infinity) or (3.08166599946613, infinity) and
    concave down on (1 - sqrt(39)/3,1 + sqrt(39)/3) or (-1.08166599946613,3.08166599946613)

    But I am not sure if it is correct. Please help
     
  2. jcsd
  3. May 15, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Mary4ever! Welcome to PF! :smile:

    (try using the X2 button just above the Reply box :wink:)

    Yes, that all looks fine. :smile:

    (I haven't checked the y coordinate calculations)

    What is worrying you about that? :confused:
     
  4. May 15, 2013 #3
    Could you please double-check it because I need to make sure everything is correct? Thank you!
     
  5. May 16, 2013 #4
    This all looks good, based on inspection of the graph and a little help from WolframAlpha.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Sketch the graphs of the functions - Calculus question
  1. Graph Sketching (Replies: 2)

  2. Sketching a graph (Replies: 2)

  3. Sketching a graph (Replies: 1)

  4. Sketching a graph (Replies: 5)

  5. Sketching a graph (Replies: 3)

Loading...