Problem 22

I solved this problem on my final exam review and came up with a working equation for 22.c), $$k(x) = 10 \sin\left(\frac \pi3x-\frac \pi6\right) - 3 $$ However, when I use $2\pi/b$ to check the period of the function, I get $6$, which is half the actual period, $12$, since the minimum value, $k(x) = -13$ has an $x$ value of $5$, and $k(x) = 7$ has an $x$ value of $11$.

Am I not accounting for the $\times2$ change in the $x$ values?

Please help, Thanks!

EDIT: Thanks G Perez for making this look better


migrated from mathematica.stackexchange.com Jun 5 '15 at 0:05

This question came from our site for users of Wolfram Mathematica.

  • $\begingroup$ Is this question about mathematics or about Mathematica software? If the former, it belongs in Mathematics. $\endgroup$ – bbgodfrey Jun 4 '15 at 23:47
  • $\begingroup$ Mathematics, sorry, I originally asked in MAthematica but it is migrated $\endgroup$ – ytpillai Jun 5 '15 at 0:14
  • $\begingroup$ Do you by chance know the answer though? :P $\endgroup$ – ytpillai Jun 5 '15 at 0:14

Try $-10 sin[(x-2) \frac{\pi}{6}] - 3$, in which case k(15) = -8.

Mathematica is convenient for quickly evaluating the function:

Table[-10 Sin[(x - 2) Pi/6] - 3, {x, 1, 15, 2}]

and yields {2, -8, -13, -8, 2, 7, 2, -8}, as expected. (The last term is the desired unknown element at x = 15.

  • $\begingroup$ Wait.....is the equation (Pi/6)x - (Pi/3)????? $\endgroup$ – ytpillai Jun 5 '15 at 14:05
  • $\begingroup$ The argument of Sin is (Pi/6)x - (Pi/3). Is that what you mean? $\endgroup$ – bbgodfrey Jun 5 '15 at 14:12
  • $\begingroup$ Yes that's what I meant thanks. I think I may have made a stupid mistake in copying my b and c answers (sin(bx-c)) to the actual equation $\endgroup$ – ytpillai Jun 5 '15 at 18:28

Your formula does not agree with the table at $x=3$, for starters. And probably at other places.

As you have noted, if the outputs $-13$ and $7$ are $6$ input units apart, and they represent a low point and high point of a sine wave, and if we assume that they represent an adjacent such pair of extrema, then the period must be $12$. (You could also deduce the period is $12$ by seeing the output $2$ occurring at inputs $1,9$, and $13$.)

So you have $$A\sin(2\pi x/12 + \varphi)+B$$

Start from there, and use other information in the table to determine $A,B,\varphi$.

  • $\begingroup$ I derived A because -3 is 10 away from the min and the max. B and c (you represented it as a greek symbol) were from plugging in x and y values, where I go values for b - c and 3b - c. $\endgroup$ – ytpillai Jun 5 '15 at 14:03
  • $\begingroup$ Sorry, I represented your B with d and 2Pi/12 as b $\endgroup$ – ytpillai Jun 5 '15 at 14:07
  • $\begingroup$ My point is that your "$\frac{\pi}{3}x$" is wrong. I have it as "$2\pi x/12$", but you could write that as "$\frac{\pi}{6}x$". $\endgroup$ – alex.jordan Jun 5 '15 at 16:52
  • $\begingroup$ Yes I realize now thanks! $\endgroup$ – ytpillai Jun 5 '15 at 18:28

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