Posted by Christina on Tuesday, August 14, 2012 at 9:09pm.
I posted this question before but accidentally typed in the wrong values! Please help, I just need to know the basic formula my book didn't provide any examples.
For the function y = 2sin(x(pi/3)) between x = 0 and x = 2pi : (6 marks)
For what value(s) of x does y have its maximum value?
For what value(s) of x does y have its minimum value?
For what value(s) of x does y=1?

Math  Reiny, Tuesday, August 14, 2012 at 11:05pm
dy/dx = 2cos(x  π/3)
at a max/min , dy/dx = 0
2cos(xπ/3) = 0
cos(xπ/3) = 0
I know that cos π/2 = 0 or cos 3π/2 = 0
so x  π/3 = π/2 or x  π/3 = 3π/2
x = 5π/6 or x = 11π/6 , (150° or 330°)
when x = 5π/6 , y = 2sin(π/2) = 2
when x = 11π/6 , y = 2sin(3π/2) = +2
so the function has a maximum value of 2 when x = 11π/3
and a minimum value of 2 when x = 5π/6
when y = 1
2sin(xπ/3) = 1
sin(xπ/3) = 1/2
I know sin 210° = 1/2 and sin 330°=1/2
which is 7π/6 or 11π/6
xπ/3 = 7π/6 or xπ/3=11π/6
x = 3π/2 or x = 13π/6
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