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Finding Horizontal tangents

  1. Feb 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Does the graph of the function
    y = 3x + 6sinx
    have any horizontal tangents in the interval 0 ≤ x ≤ 2 If so, where?

    2. Relevant equations
    Derivative rules

    3. The attempt at a solution
    I tried finding the derivative and setting it equal to zero.
    y' = 3 + 6(cosx)
    3 + 6(cosx) = 0
    but I'm not sure where to go on from here, the book only has 1 example which doesn't clear much up.
    also how do i write equations to make them look like some of the other posts i've seen(equation text lookΦ).
     
  2. jcsd
  3. Feb 13, 2016 #2

    HallsofIvy

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    You should not be taking Calculus until you are good at algebra!

    To solve an equation of the form Ax+ B= 0, first subtract B from both sides: Ax= -B.
    Now, divide both sides by A: x= -B/A.

    What do you get if you do that to your equation? What should you do now?
     
  4. Feb 13, 2016 #3

    blue_leaf77

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    Solve for ##x##.
    Use latex, https://www.physicsforums.com/help/latexhelp/.
     
  5. Feb 13, 2016 #4

    lol yea I have tried that aswell and got 2.09439
    which didn't seem similar
    [itex]

    3 +6cos(x) = 0
    [/itex]
    [itex]
    x = cos^-1(-1/2)
    [/itex]
    [itex]
    x = 2.094395102
    [/itex]
    but that wasn't which similar to the given answers of my sheet which are
    [itex]
    x = 2pi/3 , x = 4pi/3
    I'm not sure how to get to these
     
  6. Feb 13, 2016 #5
    For which x's is cos(x) equal to ##-\frac{1}{2}##? Think about the unit circle
     
  7. Feb 13, 2016 #6
    Ah i kinda see so i'm not supposed to use inverse cos, just find what value of x makes it equal to -1/2. So there was no way to solve for it without memory of it.
    is there a particular part of math that should be reviewed solve this type of problem, cartesian circle(radians and angles)?
     
  8. Feb 13, 2016 #7
    The range of the inverse cosine function is ##(0, \pi)##. So solving for x with the inverse function only gives you the solutions within that interval ##(\frac{2\pi}{3}## in this case, try multiplying it out with a calculator). But you know that the cosine function is a periodic function and it goes from 1 to -1 when x goes from 0 to ##\pi## and from -1 back to 1 when x goes from ##\pi## to ##2\pi##. So the equation ##cos(x) = k##, with ##-1 < k < 1##, will always have 2 solutions for x in ##(0, 2\pi)##. To find the second one, just note that ##cos(x) = cos(2\pi - x)##. I'd suggest you to review the geometrical interpretation of sine and cosine on the unit circle, it will help you get an intuition about why do we expect two solutions
     
    Last edited: Feb 13, 2016
  9. Feb 13, 2016 #8

    SteamKing

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    What is 2π/3 expressed as a decimal?
     
  10. Feb 13, 2016 #9

    HallsofIvy

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    Throw that calculator away and learn some mathematics!
     
    Last edited by a moderator: Feb 13, 2016
  11. Feb 13, 2016 #10

    Nugatory

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    Staff: Mentor

    Or at least put it down for a moment - and don't pick it up again until you understand what you're going to calculate and why!
     
  12. Feb 13, 2016 #11

    HallsofIvy

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    Trigonometry!
    Draw an equilateral triangle with each side equal to 1. Each angle will be equal to 60 degrees. The line from one vertex to the opposite side also bisects that angle and the opposite side. So it divides that triangle into two right triangles with acute angles 60 degrees and 30 degrees and "opposite side of length 1/2. So sin(60)= cos(30)= 1/2.
     
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