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anonymous

  • 5 years ago

Use the first derivative to find the local extrema of f and the intervals on which f is increasing and decreasing for f(x)=x+2sinx for [0,2pie]

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  1. anonymous
    • 5 years ago
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    The derivative is \[f'(x)=1+2\cos x\]This function actually tells you the slope of your original function f at any point x. When the slope is zero, the function has either stopped increasing or stopped decreasing. Either side of these points, the function is increasing or decreasing. We find the x-values for which f'(x) = 0:\[0=1+2\cos x \rightarrow \cos x =- \frac{1}{2}\]

  2. anonymous
    • 5 years ago
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    cos(x) = -1/2 when x is in the second or third quadrant. Since cos(x)=1/2 for x=60 degrees (= pi/3 radians), it follows for the second and third quadrants: x = pi - pi/3 = 2pi/3 x = pi + pi/3 = 4pi/3 will solve this equation for x in [0, 2pi).

  3. anonymous
    • 5 years ago
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    I'm putting together a plot so you can see what's going on.

  4. anonymous
    • 5 years ago
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    ok perfect! and thanks a lot! you are helping so much!

  5. anonymous
    • 5 years ago
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  6. anonymous
    • 5 years ago
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    You're only considering what's happening between the red lines (0 to 2pi).

  7. anonymous
    • 5 years ago
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    See how at x = 2pi/3 (~2.09) the function changes direction, and again at x = 4pi/3 (~4.19)?

  8. anonymous
    • 5 years ago
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    yeah those are the intervals of increasing and decreasing.

  9. anonymous
    • 5 years ago
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    Yep, but you have to use the algebraic information. If you're not using the second derivative, you may reason as follows: You can partition the interval as 0--->2pi/3--->4pi/3--->2pi

  10. anonymous
    • 5 years ago
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    Either side of the turning points, the function must be either greater or lower than the turning point, depending on the type it is. From 0 to 2pi/3, the only turning point is 2pi/3, so if the function is increasing at any two consecutive points on this interval, the function is increasing as a whole on this interval.

  11. anonymous
    • 5 years ago
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    You can test two points. Now consider what happens on the interval (2pi/3, 4pi/3)...there are no turning points in this interval, so if two consecutive function values are decreasing, then f is decreasing as a whole on this interval.

  12. anonymous
    • 5 years ago
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    You do the same for the interval, (4pi/3, 2pi)...there are no turning points, so if it's increasing for two consecutive values, it's increasing as a whole.

  13. anonymous
    • 5 years ago
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    thank you so much again! you are a god!

  14. anonymous
    • 5 years ago
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    Well, anything for The President [salute].

  15. anonymous
    • 5 years ago
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    do you think you can help me on a couple more? :)

  16. anonymous
    • 5 years ago
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    I'll try - I have to leave soon, though, that's the only thing.

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