SOLUTION: Determine the exact maximum and minimum y-values and their corresponding x-values for one period where x > 0. (For each answer, use the first occurrence for which x > 0.) f

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Question 1203489: Determine the exact maximum and minimum y-values and their corresponding x-values for one period where
x > 0.
(For each answer, use the first occurrence for which
x > 0.)
f(x)=5sin(x-5pi/6)
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

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).

Summary
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

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

Answer by ikleyn(52794)   (Show Source): You can put this solution on YOUR website!
.
Determine the exact maximum and minimum y-values and their corresponding x-values for one period where
x > 0.
(For each answer, use the first occurrence for which
x > 0.)
f(x) = 5sin(x-5pi/6)
~~~~~~~~~~~~~~~~~~~~~~~~~~


In this my post, I solve the problem for the first maximum occurrence, ONLY. 


First occurrence of the maximum is when the argument  x - 5pi/6  is  pi/2

     = .


Then  y-value is 5  and  x =  +  = write with the common denominator = 

                           =  +  =  =  = 4.18879 (rounded).


It is the ANSWER for the first maximum occurrence.


    

          Plot y =

For the first minimum occurrence, the problem is intently formulated in a way
to confuse/perplex the reader, but, due to absence the skills to write Math correctly,
the problem's composer confused/perplexed himself.

In this formulation, the meaning of the problem for the first minimum occurrence
is unclear and dark. My opinion is that the problem's formulation for the first minimum occurrence,
as it is presented/worded in the post, is INCORRECT, AMBIGOUS and SELF-CONTRADICTING.
It is why I prefer do not touch this case.


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