Consider the function f(x)=2 sin(π 2 (x−3)) +6 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 xvalues. Enter the exact answers. Amplitude: A=A= Period: P=P= Midline: y=y= The phase shift is Click for List . The vertical translation is Click for List . Hints for the maximum and minimum values of f(x)fx: The maximum value of y=sin(x)y=sinx is y=1y=1 and the corresponding xx values are x=π2x=π2 and multiples of 2π2π less than and more than this xx value. You may want to solve π2(x−3)=π2π2x−3=π2. The minimum value of y=sin(x)y=sinx is y=−1y=−1 and the corresponding xx values are x=3π2x=3π2 and multiples of 2π2π less than and more than this xx value. You may want to solve π2(x−3)=3π2π2x−3=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 xx in the interval [0, P], the minimum y-value and corresponding x-value is at: x= y=

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Consider the function f(x)=2 sin(π 2 (x−3)) +6 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 xvalues.

Enter the exact answers.

Amplitude: A=A=

Period: P=P=

Midline: y=y=

The phase shift is Click for List .

The vertical translation is Click for List .

Hints for the maximum and minimum values of f(x)fx:

The maximum value of y=sin(x)y=sinx is y=1y=1 and the corresponding xx values are x=π2x=π2 and multiples of 2π2π less than and more than this xx value. You may want to solve π2(x−3)=π2π2x−3=π2.
The minimum value of y=sin(x)y=sinx is y=−1y=−1 and the corresponding xx values are x=3π2x=3π2 and multiples of 2π2π less than and more than this xx value. You may want to solve π2(x−3)=3π2π2x−3=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: