Math question
since: \[ - 1 \leqslant \cos \left( {4x + \pi } \right) \leqslant 1\] then we can write: \[3 \leqslant - 3\cos \left( {4x + \pi } \right) + 6 \leqslant 9\] so the requested amplitude is \[A = \frac{{9 - 3}}{2} = 3\]
for period T, we can write this: \[\begin{gathered} \cos \left( {4x + \pi } \right) = \cos \left( {4x} \right)\cos \pi - \sin \left( {4x} \right)\sin \pi = \hfill \\ \hfill \\ = - \cos \left( {4x} \right) \hfill \\ \end{gathered} \] so our function becomes: \[f\left( x \right) = 3\cos \left( {4x} \right) + 6\]
now, if we make this change of variable: \[y = 4x\] then our function becomes: \[f\left( y \right) = 3\cos \left( y \right) + 6\] As you can know, the period of that function is \[T = 2\pi \] So what can you conclude?
oops..as you well know...
@vera_ewing
the amplitude of a periodic function, has to be always positive
since the cos(x) function is such that: -1 </ cos(x) </ 1, here </ stands for "less or equal to" then we can write: 3</ f(x) </ 9 so the amplitude is: A=(9-3)/2=...?
furthermore, if we make this change of variable: 4x=y, then we can rewrite your fuunction as follows: f(x)=3*cos(4x)+6, and therefore: f(y) = 3 cos(y)+6 the period of that function is T= 2*pi, so what can you conclude?
I'm sorry, I can not give the direct answer, since I have to respect the Code of Conduct
I think that option D is a wrong answer!
the period can not be 2*pi, since as I wrote before, if we use this substitution: y=4x then your function as a function of y has period = 2*pi, whereas as function of x has the period equal to: (2*pi)/4=...?
1.57
better is pi/2
now using the same substitution, we have these correspondences: x--->y/4 and therefore: pi---> pi/4 so what can you conclude about the phase shift?
The phase shift is x=pi/4 ?
the phase shift is a relative quantity, so we can assume taht it is - pi/4
that*
Oh so the answer is A!
yes!
Michele thank you so much! :)
:)
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