Results 1 to 3 of 3

Math Help - Homework # 2

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    Homework # 2

    Plot the graph of the function and from the graph, estimate the point of inflection and where the graph is concave upward and concave downward.
    Confirm your estimates analytically.
    f(x) = (x + 2)^{\frac{1}{3}}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,868
    Thanks
    745
    Hello, ^_^Engineer_Adam^_^!

    Plot the graph of the function and from the graph, estimate the point of inflection
    and where the graph is concave upward and concave downward.
    Confirm your estimates analytically.
    . . f(x) = (x + 2)^{\frac{1}{3}}
    I used an "eyeball" approach to graph it.

    We have: . y \:=\:(x + 2)^{\frac{1}{3}}\quad\Rightarrow\quad x \:=\:y^3 - 2

    This is a "sideways" cubic with its vertex at (-2,0).
    Code:
                                |
                                |         *
                                |   *
                                *
                             *  |
                           *    |
          - - - - - - - - * - - + - - - - - -
                         *      |
                       *        |
                    *           |
                *               |
          *                     |
                                |
    It appears that it is concave up for x < -2 and concave down for x > -2.


    We have: . f'(x) \:=\:\frac{1}{3}(x + 2)^{-\frac{2}{3}}

    Then: . f''(x)\:=\:-\frac{4}{9}(x + 2)^{-\frac{5}{3}} \:=\:-\frac{4}{9(x+2)^{\frac{5}{3}}}


    At x = \text{-}2\!:\:f''(-2) is undefined.

    For x < \text{-}2\!:\;f''(\text{-}3) \:=\:-\frac{4}{9(\text{-}1)^{\frac{5}{3}}} \:=\:+\frac{4}{9} . concave up: \cup

    For x > \text{-}2\!:\;f''(\text{-}1) \:=\:-\frac{4}{9(1)^{\frac{5}{3}}} \:=\:-\frac{4}{9} . concave down: \cap


    Therefore: . \begin{array}{cc}\text{concave up} & \text{on }(\text{-}\infty,\,\text{-}2) \\ \text{concave down} & \text{on }(\text{-}2,\,\infty) \end{array}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Apr 2006
    Posts
    401
    Quote Originally Posted by Soroban View Post
    Hello, ^_^Engineer_Adam^_^!

    I used an "eyeball" approach to graph it.

    We have: . x + 2)^{\frac{1}{3}}\quad\Rightarrow\quad x \:=\:y^3 - 2" alt="y \:=\x + 2)^{\frac{1}{3}}\quad\Rightarrow\quad x \:=\:y^3 - 2" />

    This is a "sideways" cubic with its vertex at (-2,0).
    Code:
                                |
                                |         *
                                |   *
                                *
                             *  |
                           *    |
          - - - - - - - - * - - + - - - - - -
                         *      |
                       *        |
                    *           |
                *               |
          *                     |
                                |
    It appears that it is concave up for x < -2 and concave down for x > -2.


    We have: . f'(x) \:=\:\frac{1}{3}(x + 2)^{-\frac{2}{3}}

    Then: . f''(x)\:=\:-\frac{4}{9}(x + 2)^{-\frac{5}{3}} \:=\:-\frac{4}{9(x+2)^{\frac{5}{3}}}


    At x = \text{-}2\!:\:f''(-2) is undefined.

    For x < \text{-}2\!:\;f''(\text{-}3) \:=\:-\frac{4}{9(\text{-}1)^{\frac{5}{3}}} \:=\:+\frac{4}{9} . concave up: \cup

    For x > \text{-}2\!:\;f''(\text{-}1) \:=\:-\frac{4}{9(1)^{\frac{5}{3}}} \:=\:-\frac{4}{9} . concave down: \cap


    Therefore: . \begin{array}{cc}\text{concave up} & \text{on }(\text{-}\infty,\,\text{-}2) \\ \text{concave down} & \text{on }(\text{-}2,\,\infty) \end{array}

    Your second derivative should be:

    -2/(9(x + 2)^5/3)

    But, it did not change the results;

    f''(-1) = -2/9
    f''(-3) = 2/9,
    f''(-2) still undef

    Thus, we have a change in concavity at x = -2

    And CD from -2 to inf; CU from -inf to -2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. homework help
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: August 31st 2009, 05:36 AM
  2. need help with homework
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: July 20th 2009, 09:34 AM
  3. homework help
    Posted in the Algebra Forum
    Replies: 2
    Last Post: February 10th 2009, 08:12 PM
  4. Homework Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 2nd 2007, 02:09 PM
  5. Homework Help
    Posted in the Math Topics Forum
    Replies: 9
    Last Post: October 14th 2007, 04:00 PM

Search Tags


/mathhelpforum @mathhelpforum