For the function,y=-2sin( x-π/3),between x=0 and x=2π,for what value(s)of x does y have it max value? For what value(s) of x does y have it min value?For what value(s) of x does y = 0?

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`y=-2sin( x-pi/3)`

Here y depends on the `sin(x-pi/3)` term.

The max value of `sin(x-pi/3) = 1`

The min value of `sin(x-pi/3) = -1`

 

Since `y=-2sin( x-pi/3)` ; y would have a maximum when `sin(x-pi/3) = -1`

`sin(x-pi/3) = -1`

`sin(x-pi/3) = sin 3pi/2`

              `x = 3pi/2+pi/3`

                 `= 11pi/6`

 

So we have a maximum between `x in (0,2pi)` at `x = 11pi/6`

Maximum y = -2*-1 = 2

 

y will be minimum when `sin(x-pi/3) = 1`

`sin(x-pi/3) = 1`

`sin(x-pi/3) = sin pi/2`

             `x = pi/2+pi/3`

                `= 5pi/6`

 

So minimum of y will be at` x = 5pi/6`

minimum y = -2*1 = -2

 

y = 0 when `sin(x-pi/3) = 0`

`sin(x-pi/3) = 0`

`sin(x-pi/3) = sin 0 `

              `x = pi/3`

 

So y = 0 at `x = pi/3`

 

Note:

There is no need to go for general solutions for sine since `x in (0,2pi)`

Sources:

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