Hi everyone, I have this pic
I found f(x)= 3 sin ([FONT=MathJax_Math]
π[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Main]2[/FONT]*(x-1)) -1 but the book says c=3, how is it correct?
If we know it is a sine wave of the form
f(x) = a sin[ k (x - c) ] + d
Short Version [but make sure you see below for Note(1) & Note (2)]:
At x = c, the sine is zero and f(x) = d, i.e. c is the the first d crossing (not the first zero crossing). Thus, in this problem c can be taken as 1 but that will affect some of the rest of your values.
Long Version:
then the
max = |a| + d
min = -|a| + d
So
|a| = (max - min) / 2
and, going through the math,
d = (max + min) / 2
In this case |a| = [2 - (-4)] / 2 = 3 and d = [2 + (-4)] / 2 = -1
At x = c (or \(\displaystyle \pi\) multiples), the sine is zero and f(x) = d, i.e. c is a d crossing (not the first zero crossing). Thus, in this problem we can take c as 1, 3, 5, .... [see Notes below]
The distance between two d crossings is half a period, that is the argument increases by \(\displaystyle \pi\). So, letting F be a d crossing and S be the next d crossing we have
k (S - c) - k (F - c) = \(\displaystyle \pi\)
or, doing the math
k = \(\displaystyle \frac{\pi}{S - F}\)
or, in this problem k = \(\displaystyle \frac{\pi}{S - F} = \frac{\pi}{2}\)
NOTE (1): c is not unique. Since the sine has zeros at arguments of n\(\displaystyle \pi\), we can use any c which satisfies our c above (call that c
0) such that
k (x - c
0) = k (x - c) + n \(\displaystyle \pi\)
[the n \(\displaystyle \pi\) is because we haven't yet set the sign of a] or
c = c
0 + n \(\displaystyle \pi\) / k
In the problem here, since k = \(\displaystyle \frac{\pi}{2}\) we have
c = 1 + 2 n
Letting n=1 we could also use c = 3.
Note (2): Notice that I never did find out what a was although I did find the magnitude of a. I'll leave that for the student. Hint: It is related to the 1 (first d crossing) or 3 (second d crossing) type choice, i.e. look at the derivative.
