Respuesta :

Answer:

Problem 13) [tex]f(x)=4\,sin(\frac{1}{2} x+\frac{2}{3}\pi )-2[/tex]

Problem 14) [tex]f(x)=cotan(x+\frac{1}{3} \pi)+2[/tex]

Step-by-step explanation:

Recall how transformations affect the graph of the sine function, and how such is conveyed into the parameters A, B, C, and D that could be included in the general form of the function:

[tex]f(x)=A\,sin(Bx+C)+D[/tex]

where the Amplitude of the transformed sine function is the absolute value of the multiplicative parameter A:

Amplitude = [tex]|A|[/tex]

The period is (which for sin(x) is [tex]2\pi[/tex]) is modified by the parameter B in the following manner:

Period = [tex]\frac{2\pi}{B}[/tex]

Where the phase shift is introduced as:

Phase shift = [tex]-\frac{C}{B}[/tex].

and finally any vertical shift is included by the constant D (positive means shift upwards in D many units, and negative means shift downwards D units)

Therefore, to have a sine function with the requested characteristics, we work on the value of the parameters A, B, C, and D one at a time:

1) Amplitude = [tex]|A|=4[/tex] then we use parameter A = 4

[tex]f(x)=4\,sin(Bx+C)+D[/tex]

2) Period [tex]4\pi[/tex], then we work on the parameter B:

Period = [tex]\frac{2\pi}{B}[/tex]

[tex]4\pi=\frac{2\pi}{B}\\B*4\pi=2\pi\\B=\frac{2\pi}{4\pi} \\B=\frac{1}{2}[/tex] which transforms the function into:

[tex]f(x)=4\,sin(\frac{1}{2} x+C)+D[/tex]

3) phase-shift = [tex]-\frac{4}{3} \pi[/tex]

Then knowing that B=[tex]\frac{1}{2}[/tex], we work on the value of parameter C:

Phase shift = [tex]-\frac{C}{B}[/tex]

[tex]-\frac{4}{3} \pi=-\frac{C}{B} \\-\frac{4}{3} \pi=-\frac{C}{ \frac{1}{2} }\\-\frac{4}{3}* \frac{1}{2} \pi=-C\\C=\frac{2}{3} \pi[/tex]

Therefore the function gets transformed into:

[tex]f(x)=4\,sin(\frac{1}{2} x+\frac{2}{3}\pi )+D[/tex]

4) and finally the vertical shift of negative two units, that gives us the value D = -2

The complete transformed function becomes:

[tex]f(x)=4\,sin(\frac{1}{2} x+\frac{2}{3}\pi )-2[/tex]

Now for problem 14, recall that the cotangent function is the reciprocal of the tangent function, therefore, their periodicity is the same: [tex]\pi[/tex]

since you are asked for a cotangent function of period [tex]\pi[/tex] as well, there is no multiplication parameter "B" needed (so we keep it unchanged - equal to one). B = 1

Then for the phase-shift which we want it to be [tex]-\frac{1}{3} \pi[/tex], we set the condition:

[tex]-\frac{1}{3} \pi=-\frac{C}{B} \\-\frac{1}{3} \pi=-\frac{C}{1}\\-\frac{1}{3} \pi=-C\\C=\frac{1}{3} \pi[/tex]

And insert such in the cotangent general form:

[tex]f(x)=cotan(x+\frac{1}{3} \pi)+D[/tex]

and finally include the desired vertical shift of 2 units:

[tex]f(x)=cotan(x+\frac{1}{3} \pi)+2[/tex]

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