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Thread: Point of Inflexion

  1. #1
    Junior Member
    Nov 2008
    Far away.

    Smile Point of Inflexion

    I have tried to solve this question about the function f(x) = cos x + sin x
    But it was a bit difficult for me.
    The question is:
    Consider the function f(x) = cos x + sin x

    (i) Show that f(– p/4) = 0.
    (ii) Find in terms of p, the smallest positive value of x which satisfies f(x) = 0.

    b) The diagram shows the graph of y = ex (cos x + sin x), – 2 £ x £ 3. The graph has a maximum turning point at C(a, b) and a point of inflexion at D.
    (I have attached the diagram)

    (i) find dy/dx
    (ii) Find the exact value of a and of b.
    (iii) Show that at D, y = sqrt2 e^ p/4

    Thank you very much for your help!
    Best wishes,

    Attached Thumbnails Attached Thumbnails Point of Inflexion-ko.bmp  
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  2. #2
    Senior Member
    Dec 2008
    a) We may remember that if \theta measures the arc distance from a point on the unit circle counterclockwise from (x,y)=(1,0), then \cos \theta gives the x-coordinate of that point and \sin \theta gives the y-coordinate.

    Therefore, \cos \theta+\sin \theta = 0 when \cos \theta=-\sin \theta, i.e, when the y-coordinate of the point on the circle is the negative of its x-coordinate. What fraction of the total arc length, 2\pi, do these points give?

    b) Points of inflection are found where y''=0 and y' differs in sign on both sides next to the point. Extrema are found at critical points: in this case, as y is differentiable and has no boundary points, at points where y'=0. To differentiate

    y=e^x(\cos x + \sin x),

    we apply the Product Rule:

    \frac{dy}{dx}=\left(\frac{d}{dx}e^x\right)(\cos x+\sin x)+e^x\left(\frac{d}{dx}(\cos x+\sin x)\right).

    When is y'=0? When is y''=0? Hint: e^x is always positive.

    Hope this helps!
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