Results 1 to 4 of 4

Math Help - Horizontal Tangent

  1. #1
    Newbie
    Joined
    Mar 2009
    From
    USA
    Posts
    21

    Horizontal Tangent

    1. f(x)= x+2cosx

    2. f(x)= x+2sinx

    3. f(x)= e^x (sinx)

    Regarding problem number 1, I know that the derivative is 1-2sinx, then when I set it equal to zero I get sinx=1/2. Therefor it would be pi/6, but the answer key says that the answer is 2npi + pi/6, 2npi + 5pi/6. I need some clarification on how to reach that conclusion.

    On problem 2, I get the answer cosx= -1/2. The answer key says that the final answer is (2n +1)pi +/- pi/3. Again I am quite sure how to reach that conclusion or final answer.

    Same problem on 3. tanx= -1: What would my next step be to solve the problem.

    Any feedback or help would be appreciated. These problems have been bugging me all day.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Gitano View Post
    1. f(x)= x+2cosx

    2. f(x)= x+2sinx

    3. f(x)= e^x (sinx)

    Regarding problem number 1, I know that the derivative is 1-2sinx, then when I set it equal to zero I get sinx=1/2. Therefor it would be pi/6, but the answer key says that the answer is 2npi + pi/6, 2npi + 5pi/6. I need some clarification on how to reach that conclusion.

    This is just trigonometry: \color{red}\sin x=\sin (\pi-x)\, \mbox{so}\,\sin x=\frac{1}{2}\Longrightarrow x=\frac{\pi}{6}+2n\pi
    \color{red}\mbox{or}\,\, x=\pi-\frac{\pi}{6}+2n\pi=\frac{5\pi}{6}+2n\pi

    Remember that both basic trigonometry functions \sin x\, \mbox{and} \cos x have period =2\pi

    For the next two problems the issue is the same, only that \tan x has period \pi

    Tonio

    On problem 2, I get the answer cosx= -1/2. The answer key says that the final answer is (2n +1)pi +/- pi/3. Again I am quite sure how to reach that conclusion or final answer.

    Same problem on 3. tanx= -1: What would my next step be to solve the problem.

    Any feedback or help would be appreciated. These problems have been bugging me all day.
    ..
    Last edited by tonio; October 19th 2009 at 10:11 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    From
    USA
    Posts
    21
    Thanks for the clarification, appreciate it.

    But I don't still don't quite understand problem 3. I get tan x = -1 and the answer key says that x= npi + pi/4. Tangent is negative in quadrants 2 and 4, so I don't understand how it is just pi/4.
    Last edited by Gitano; October 19th 2009 at 10:53 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Gitano View Post
    Thanks for the clarification, appreciate it.

    But I don't still don't quite understand problem 3. I get tan x = -1 and the answer key says that x= npi + pi/4. Tangent is negative in quadrants 2 and 4, so I don't understand how it is just pi/4.

    I don't understand it either: \tan x = -1 \Longrightarrow x=\frac{3\pi}{4}+n\pi , n\in\mathbb{N}

    It must be a mistake, I suppose.

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. horizontal tangent, set it to zero, now what?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 11th 2010, 10:19 PM
  2. Horizontal tangent
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 11th 2010, 03:41 PM
  3. horizontal tangent
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 9th 2009, 08:34 PM
  4. horizontal tangent
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 26th 2009, 09:49 PM
  5. horizontal tangent
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 28th 2008, 11:30 PM

Search Tags


/mathhelpforum @mathhelpforum