Results 1 to 6 of 6

Thread: Inflection Points

  1. #1
    Senior Member topher0805's Avatar
    Joined
    Jan 2008
    From
    Vancouver
    Posts
    336

    Inflection Points

    I already know my first derivative to be:

    $\displaystyle f'(x) = 16\cos {x}(-\sin {x} - 1)$

    So I solved for my second derivative and got:

    $\displaystyle -16\sin {x}(-\sin {x} - 1) + 16\cos {x}(-\cos {x})$

    $\displaystyle = 16\sin {x}(\sin {x} + 1) - 16\cos^2 {x}$

    I'm looking for where $\displaystyle f$ is concave up and concave down, and I figured that inflection points were a good way to find that. So, I set my equation equal to zero:

    $\displaystyle 16\sin {x}(\sin {x} + 1) - 16\cos^2 {x} = 0$

    $\displaystyle 16\sin {x}(\sin {x} + 1) = 16\cos^2 {x}$

    $\displaystyle \sin {x}(\sin {x} + 1) = \cos^2 {x}$

    I am completely stuck here. How do I solve for $\displaystyle x$ now?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,410
    Thanks
    1
    Assuming everything you've done so far is correct:
    $\displaystyle sinx(sinx+1) = cos^{2}x$

    $\displaystyle sin^{2}x + sinx = 1 - sin^{2}x$

    $\displaystyle 2sin^{2}x + sinx - 1 = 0$

    Solve the quadratic.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2008
    Posts
    148
    The sign of the second derivative will indicate which direction the function opens: positive second derivative means concave up, negative second derivative means concave down. So those intervals on which the function has a positive second derivative are the intervals in which the function is concave up, vice versa for points where the function has a negative second derivative.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member topher0805's Avatar
    Joined
    Jan 2008
    From
    Vancouver
    Posts
    336
    Quote Originally Posted by o_O View Post
    Assuming everything you've done so far is correct:
    $\displaystyle sinx(sinx+1) = cos^{2}x$

    $\displaystyle sin^{2}x + sinx = 1 - sin^{2}x$

    $\displaystyle 2sin^{2}x + sinx - 1 = 0$

    Solve the quadratic.
    How do you solve the quadratic when it involves trig?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2008
    Posts
    148
    Treat it like a polynomial in $\displaystyle \sin(x)$. If you like it better say $\displaystyle \sin(x) = y$.

    Then your quadratic becomes $\displaystyle 2y^2+y-1 = 0$

    once you solve for the possible values of $\displaystyle y$ substitue $\displaystyle \sin(x)$ back in and solve for $\displaystyle x$ using $\displaystyle \arcsin$ or some other technique.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member topher0805's Avatar
    Joined
    Jan 2008
    From
    Vancouver
    Posts
    336
    Quote Originally Posted by iknowone View Post
    Treat it like a polynomial in $\displaystyle \sin(x)$. If you like it better say $\displaystyle \sin(x) = y$.

    Then your quadratic becomes $\displaystyle 2y^2+y-1 = 0$

    once you solve for the possible values of $\displaystyle y$ substitue $\displaystyle \sin(x)$ back in and solve for $\displaystyle x$ using $\displaystyle \arcsin$ or some other technique.
    So I would have:

    $\displaystyle
    (2y-1)(y+1)=0$

    $\displaystyle (2\sin {x} - 1)(\sin {x} + 1) = 0$

    So x is 0 when $\displaystyle \sin {x} = -1, \frac {1}{2}$

    Thanks guys I appreciate it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Aug 24th 2011, 11:35 AM
  2. Replies: 16
    Last Post: Jun 10th 2011, 06:49 AM
  3. Replies: 1
    Last Post: Apr 20th 2010, 07:36 PM
  4. Replies: 2
    Last Post: Oct 29th 2009, 08:02 PM
  5. Replies: 5
    Last Post: Feb 27th 2009, 07:48 PM

Search Tags


/mathhelpforum @mathhelpforum