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Mathematics 17 Online
OpenStudy (anonymous):

hellpppp

OpenStudy (anonymous):

OpenStudy (anonymous):

isn't that 1?

OpenStudy (bahrom7893):

f'(x) = 2Cosx - 2Sin2x = 0

OpenStudy (bahrom7893):

solve for x: 2Cosx - 2Sin(2x) = 0

OpenStudy (bahrom7893):

x = (1/2) pi (4*n-1), where n is an integer

OpenStudy (bahrom7893):

the only solution in the given interval is: x = 3pi/2

OpenStudy (bahrom7893):

So plug each endpoint and the solution into the equation: x=0 y= 2Sin0 + Cos(2*0) = 0 + 1 = 1

OpenStudy (bahrom7893):

x = 3pi/2 y = 2Sin(3pi/2) + Cos(3pi) = -3

OpenStudy (bahrom7893):

And finally, 2pi: x = 2pi y = 2Sin(2pi) + Cos(4pi) = 1

OpenStudy (bahrom7893):

Thus the maximum of y is equal to 1, which is none of the above, which is option 1.

OpenStudy (anonymous):

thank you so much! what would be the absolute minimun?

OpenStudy (bahrom7893):

absolute minimum is y=-3, which occurs at x = 3pi/2 (assuming the interval's the same)

OpenStudy (anonymous):

oh ok and to find all the solutions for sec4x=2 , i solved it and i got pi/12 but i dont know how to find the other solutions

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