A minimum value of a sinusoidal Function is at #(pi/4, 3)#. The nearest maximum value to the right of this point is at #((7pi)/12, 7)#. What is the equation of this function?

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Answer 1 · 2016-12-30T14:13:57

#y(x) = 2sin(3x-(5pi)/4)+5#

Let us write the generic sinusoidal function as:

#y(x) = Asin(alphax+beta)+B#

First we note that the value for the minimum is #y_m=3# while the maximum is for #y_M=7#. From this we may derive:

#-A+B = 3#
#A+B = 7#

and therefore:

#A=2# and #B=5#

Then we note that a minimum of #sinx# occurs for #x=-pi/2# and the next maximum for #x=pi/2#, so we can derive:

#alpha x_m + beta = -pi/2#
#alpha x_M + beta = pi/2#

or:

#alpha pi/4 + beta = -pi/2#

#alpha(7pi)/12+beta = pi/2#

subtracting the first equation from the second:

#alpha((7pi)/12-pi/4) = pi#

so

#alpha=1/(7/12-1/4) = 1/(7/12-3/12) = 12/4=3#

and

#beta = -pi/2-3/4pi = -5/4pi#

Thus the sinusoidal function we are searching is:

#y(x) = 2sin(3x-(5pi)/4)+5#

graph{2sin(3x-(5pi)/4)+5 [-2.396, 2.604, 2, 8]}