Consider the function f(x)=2 \sin(\pi2(x−3))+4. State the amplitude A

glamrockqueen7

glamrockqueen7

Answered question

2021-10-14

Consider the function f(x)=2sin(π2(x3))+4.
State the amplitude A, period P, and midline. State the phase shift and vertical translation. In the full period [0, P], state the maximum and minimum y-values and their corresponding x-values.
Hints for the maximum and minimum values of f(x):
The maximum value of y=sin(x) is y=1 and the corresponding x values are x=π2 and multiples of 2π less than and more than this x value. You may want to solve π2(x3)=π2.
The minimum value of y=sin(x) is y=1 and the corresponding x values are x=3π2 and multiples of 2π less than and more than this x value. You may want to solve π2(x3)=3π2.
If you get a value for x that is less than 0, you could add multiples of P to get into the next cycles.
If you get a value for x that is more than P, you could subtract multiples of P to get into the previous cycles.
For x in the interval [0, P], the maximum y-value and corresponding x-value is at:
x=
y=
For x in the interval [0, P], the minimum y-value and corresponding x-value is at:
x=
y=

Answer & Explanation

Benedict

Benedict

Skilled2021-10-15Added 108 answers

Step 1 Consider a given function
f(x)=2sin[π2(x3)+4].
State the amplitude A, period P and mid-line.
State the phase shift and vertical translation.
State the minimum and maximum y-values and their corrosponding x-values in the interval [0,P].
Step 2
a) Consider a given function
f(x)=2sin[π2(x3)]+4
The amplitude of the function f is 2. Hence, A=2.
The period of the function f is 2π(π2)=4. Hence, T=4.
The mid-line of the trigonometric function is a point where the function attain its maximum values.
Now,
π2(x3)=(2n1)π2
x3=2n1
x=2n+2 for all nZ
Hence, the mid-line of the function f is x=2n+2 for all nZ
Step 3
b) Consider a given function
f(x)=2sin[π2(x3)]+4
The function f is shifted right side by 3 units. Hence, its phase shift is 3.
The function f is shifted vertically upward by 4 units. Hence, its vertically shift is 4 units.
Step 4
c)
The range of the trigonometric function is [1,1].
It implies that 1sin(x)1.
Hence,
1sin(π2(x3))1
22sin(π2(x3))2
2+42sin(π2(x3))+42+4
2f(x)6
Hence, minimum value of the function f is 2 and the maximum value of the function f is 6.
It implies that fmin=2 and fmax=6.
Step 5
Further,
Maximum attain at x=2n+2 for all nZ.
Hence, the point of maxima is x=0,2,4.
sin(π2(x3))=0
π2(x3)=nπ
x3=2n
x=2n+3 for all nZ
Hence, the point of minima is 1,3.

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