anonymous
  • anonymous
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]
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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:\[0=1+2\cos x \rightarrow \cos x =- \frac{1}{2}\]
anonymous
  • anonymous
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).
anonymous
  • anonymous
I'm putting together a plot so you can see what's going on.

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anonymous
  • anonymous
ok perfect! and thanks a lot! you are helping so much!
anonymous
  • anonymous
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anonymous
  • anonymous
You're only considering what's happening between the red lines (0 to 2pi).
anonymous
  • anonymous
See how at x = 2pi/3 (~2.09) the function changes direction, and again at x = 4pi/3 (~4.19)?
anonymous
  • anonymous
yeah those are the intervals of increasing and decreasing.
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
thank you so much again! you are a god!
anonymous
  • anonymous
Well, anything for The President [salute].
anonymous
  • anonymous
do you think you can help me on a couple more? :)
anonymous
  • anonymous
I'll try - I have to leave soon, though, that's the only thing.

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