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Find all points on graph of f(x)=2sinx+sin^2x where slope = 0

  1. Jul 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Find all points on the graph of the function

    [tex]f(x)=2sinx+sin^2x[/tex]

    at which the tangent line is horizontal.


    2. Relevant equations
    - Power rule
    - Chain rule
    - Product rule?



    3. The attempt at a solution
    So I want all points at which the tangent line to this function has a slope of zero.

    [tex]f(x)=2sinx+sin^2x[/tex]

    [tex]f'(x)=2cosx+(sinx)^2[/tex]

    [tex]f'(x)=2cosx+2sinx*\frac{d}{dx}sinx[/tex]

    [tex]f'(x)=2cosx+2sinx*cosx[/tex]

    [tex]f'(x)=cosx(2+2sinx)[/tex]

    ...?

    How do I go from here? I know to set that last equation equal to zero and find all the solutions, but just how do I find those solutions?

    The solution from the book says:
    [tex](\frac{\pi}{2}+2n\pi,3),(\frac{3\pi}{2}+2n\pi,-1)[/tex] where n an integer
     
  2. jcsd
  3. Jul 27, 2008 #2

    rock.freak667

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    Homework Helper

    Slope=0 means f'(x)=0

    then you use the general rule that if AB=0 then,A=0,B=0
     
  4. Jul 27, 2008 #3
    Right, I know.. but I don't see how to arrive at those values given in the solution. Is my calculator required here?
     
  5. Jul 27, 2008 #4
    First, think back to algebra when you had those equations like (x-4)(x-2)=0, and you solved them by setting up two possibilities: either x-4=0, or x-2=0. You can use the same principle here.

    If [tex]cosx=0[/tex], or if [tex]2+2sinx=0[/tex], then [tex]f'(x)=0[/tex].

    Now simply find all the points where this can happen. You don't need a calculator - simple trigonometry can help. Where is [tex]cosx=0[/tex]? And where is [tex]sinx=-1[/tex]?
     
  6. Jul 27, 2008 #5

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi illjazz! :smile:

    (have a pi: π and a squared: ² :smile:)

    I think you're misunderstanding the notation in the solution.

    sin(π/2 + 2nπ) = sin(π/2) = 1, and so f(π/2 + 2nπ) = 3.

    sin(3π/2 + 2nπ) = sin(3π/2) = … ?, and so f(3π/2 + 2nπ) = … ?. :smile:
     
  7. Jul 27, 2008 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, it is not. I know what you mean but this is terrible notation. You mean
    [tex]f'(x)= 2 cosx+ \frac{d}{dx} (sin x)^2[/tex]


    Okay, now that is correct.

    But why did you do this? Just solve 2cos x+ 2 sin x cos x= 0. That is the same as
    2 cos x (1+ 2 sin x)= 0 which means either cos x= 0 or 1+ 2 sin x= 0. That is easy to solve.

     
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