Question 1203489
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The min and max of y = sin(x) are y = -1 and y = 1 respectively.
This would apply even if the input x is replaced with any expression such as x - 5pi/6


Sticking a 5 out front will vertically stretch the graph. 
This would mean the min and max of y = 5*sin(x-5pi/6) are y = -5 and y = 5 respectively. 
Use a graphing tool like a TI83, Desmos, or GeoGebra to confirm this claim. 


Let's replace y with -5 and solve for x.
y = 5*sin(x-5pi/6)
-5 = 5*sin(x-5pi/6)
-5/5 = sin(x-5pi/6)
-1 = sin(x-5pi/6)
x-5pi/6 = arcsin(-1) or x-5pi/6 = pi - arcsin(-1)
x-5pi/6 = 3pi/2 or x-5pi/6 = pi - 3pi/2
x-5pi/6 = 3pi/2 or x-5pi/6 = 2pi/2 - 3pi/2
x-5pi/6 = 3pi/2 or x-5pi/6 = -pi/2
x = 3pi/2+5pi/6 or x = -pi/2+5pi/6
x = 9pi/6+5pi/6 or x = -3pi/6+5pi/6
x = 14pi/6 or x = 2pi/6
x = 7pi/3 or x = pi/3


Because 1 < 7, it leads to pi < 7pi and further pi/3 < 7pi/3
pi/3 is the smaller of the two values.


This is the first occurrence when we reach the min y = -5 such that x > 0.


To find when the max y = 5 first occurs, plug in y = 5 and follow similar steps as shown above.
After doing so, you should find that x = 4pi/3 is the first x value to reach the max y = 5 when x > 0.


The graphing tools I mentioned earlier can be used to confirm the answers. 
Various online calculators can be helpful as well. 
I recommend that you do the actual scratch work yourself and use the calculators to check your answers (or else you won't be ready for the exams later).


<font color=red size=4>Summary</font>
Min and max are y = -5 and y = 5
First min occurs when x = pi/3
First max occurs when x = 4pi/3
In other words, the location of the first min is (pi/3, -5) when x > 0.
The location of the first max is (4pi/3, 5) when x > 0.


Graph
{{{drawing(400,400,-7,7,-7,7,
graph(400,400,-7,7,-7,7,-1000,5*sin(x-5pi/6)),

circle(1.04719755,-5,0.1),circle(1.04719755,-5,0.12),circle(1.04719755,-5,0.14),circle(1.04719755,-5,0.16),circle(4.18879020,5,0.1),circle(4.18879020,5,0.12),circle(4.18879020,5,0.14),circle(4.18879020,5,0.16),

locate(1.2472,-5.2,"A"),locate(4.58879,5.4,"B")

)
}}}
A = first local min = (pi/3, -5) when focusing on the interval x > 0
B = first local max = (4pi/3, 5) when focusing on the interval x > 0
pi/3 = 1.04719755 approximately
4pi/3 = 4.18879020 approximately

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