Results 1 to 4 of 4

Math Help - Points of Inflexion on a Trigonometric Graph

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    45

    Points of Inflexion on a Trigonometric Graph

    I'm having some trouble with this question, I was hoping someone could help me out. The question asks;

    Find all points of inflexion on the curve y = 3\cos(2x + \frac{\pi}{4}).

    Show all your working, please.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Ok

    Quote Originally Posted by Flay View Post
    I'm having some trouble with this question, I was hoping someone could help me out. The question asks;

    Find all points of inflexion on the curve y = 3\cos(2x + \frac{\pi}{4}).

    Show all your working, please.
    f'(x)=-6sin\bigg(2x+\frac{\pi}{4}\bigg)... f''(x)=-12cos\bigg(2x+\frac{\pi}{4}\bigg)...no intervals so I am going to assume it is [0,2\pi] the solutions are x=.3926,x=1.9635,x=3.534,x=5.105...now using those set up test intervals and test an element of each in f''(x) where it changes signs there is an inflection point
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2008
    Posts
    45
    Sorry, forgot to mention the interval. You were right in assuming [0, 2\pi]. What I don't get about this question, though, is exactly how you solve f''(x)=-12cos\bigg(2x+\frac{\pi}{4}\bigg) to get those values. Could you explain, please?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641

    Sure

    Quote Originally Posted by Flay View Post
    Sorry, forgot to mention the interval. You were right in assuming [0, 2\pi]. What I don't get about this question, though, is exactly how you solve f''(x)=-12cos\bigg(2x+\frac{\pi}{4}\bigg) to get those values. Could you explain, please?
    ...there are two ways you need to solve this -12cos\bigg(2x+\frac{\pi}{4}\bigg)=0...first way cos\bigg(2x+\frac{\pi}{4}\bigg)=0,2x+\frac{\pi}{4}  =\bigg(arccos(0)=\frac{\pi}{2}\bigg),x=\frac{\pi}{  8} then using the trig identites find the other values(e.g. add \frac{\pi}{2})...secondly graph it on your calculator and find the points of intersection between [tex][0,2\pi]]/math]
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Point of Inflexion
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 19th 2009, 03:23 AM
  2. Inflexion point
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 5th 2009, 08:05 PM
  3. maxima, minima, points of inflexion
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 6th 2008, 06:46 PM
  4. Point of inflexion
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 8th 2007, 01:06 PM
  5. Dot inflexion
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: August 18th 2007, 08:43 AM

Search Tags


/mathhelpforum @mathhelpforum