Question

How do you determine a maximum and minimum value without graphing. I was given an equation..could that help?

Answer 1

Calculus is the easiest way, though you said in an earlier question that you haven't taken it yet. Perhaps give your equation, since there are non-calculus techniques that can be used for certain types of equations, like sinusoidal curves, that I could show you.

Answer 2

alright my equation follows:

\[P(t) = 5000\sin(30(t-7)) + 8000

where t is the number of months after New Years

Answer 3

ignore that slash after 5000

Answer 4

Ah a quick note, if you want openstudy to register what you write as an equation, you also have to end it with \]. And you can leave it as \sin, rather than sin, it makes it look a little nicer.

To answer your question, do you know the range of values that sine can take?

Answer 5

Observe using \sin \[\sin x\] versus just sin \[sin x\]

Answer 6

\[P(t) = 5000\sin (30(t-7)) + 8000\]

Answer 7

what do you mean by the range of values sin can take?

Answer 8

Okay, if you consider \[y = \sin x\] where x can be any real number, what are all possible values that y can be?

Answer 9

y = -1, 0, 1

Answer 10

Even more than that, sine can take on ALL values between -1 and 1 (inclusive). If you consider what a general sine curve looks like, you can see this for yourself.

Now, since sin x can be anything from -1 to 1 and will ALWAYS take on those values, if you want to find the maxima and minima of a sinusoidal curve, say \[f(x)=A\sin(Bx-C)+D\] Then the smallest value sin can be is -1 and the largest it can be is 1. Thus, if you plug in 1 for sin, you will get the maxima: \[f(x)=A(1)+D=A+D\] and plugging in -1 for sin gives the minima: \[f(x)=A(-1)+D=-A+D\] This works for all sine and cosine curves.

Answer 11

thankkksss :)
but for the value (t) what do i plug in
12 months?

Answer 12

Well, if you want the maximum of this curve, you would just need to find points where \[\sin(30(t-7))=1\] This will equal 1 an infinite number of times and a little trig will show that it equals 1 if \[30(t-7)=\frac{\pi}{2}+2\pi k\] for ANY integer value of k (..., -3, -2, -1, 0, 1, 2, 3, ...) If you don't care which maximum point you have, just set k=0. Then you have \[30(t-7)=\frac{\pi}{2}\] which you can easily solve for t. Unfortunately, as you will often see with trigonometry, the answer isn't a simple number like 12.

Answer 13

you seriously explain things far better than my math teacher
thanks dude (:
well i wont be focusing on pi till next year
although i figured the answers to this question using the steps you showed. Thanks again

Answer 14

No problem :) I love teaching and I'm always glad to help when people need it.

Answer 15

I'd be surprised if he was working in radians...I'm betting degrees.

Answer 16

Ah yes, that is true, if you replace the 30 with pi/6, then you get 10 months. I forgot that you are using degrees in your answers and combined your degrees with my radians.

Answer 17

Aha yeah i got 10 months for my maxima and for my minima i got 4 months. So far so good. I got 2 more questions to go until starting on my chemistry.