What if the questions gives no domain?
Separate question:
Why does the CAS have so much difficulty solving maximum and minimum value of circular functions? My friend said I need to do some by hand because the solution is general but what exactly do I do by hand?
Okay, I think you might have some confusion about what circular functions exactly /are/. Circular functions are periodic - they have a period, that means after that period, every time the function repeats. "Yes keltingmeith, I know that, stop being so condescending. I know how to solve for the period, it's 2pi/n" - is probably what you're thinking, but let's take a step back and think about what this actually /means/ by looking at some graphs:
Let's look at the graph of y=sin(pi*(x-0.5))You're probably familiar with this graph. You've sketched it a countless amount of times. What does this have to do with anything? Well, let's focus on the periodicity - the periodicity is 2, which means the graph repeats ever 2 units. Notice how the first maximum (after x=0) for the graph is 1? That means that there's going to be another maximum every time you add 2 - so there'll be another one at 3, another one at 5, another one at 7, etc. This graph has a maximum EVERY TIME x=an odd number. Well, that would mean that we have the solution:
Oh, but wait, those aren't all the odd numbers. There's also:
Phew, I'm running out of space here... And I haven't even started on the negative odd numbers yet!! We need another system - we need some equation that tells us whether the number is odd or not. So, how do you indicate whether a number is odd or not? This is hopefully something you encountered earlier in methods - for a number z to be odd, then it must be equal to some number doubled, plus 1. Or:
To figure out what values z takes on, all you need to do is substitute in different numbers. For example:
+1=1\\<br />n = 1\implies z=2(1)+1=3\\<br />n=2\implies z=2(2)+1=5\\<br />n=3\implies z=2(3)+1=7)
And you get the idea. THIS is what we mean by a general solution - we want a solution that tells you where ALL of the maximum points are, and it's at the points x=2n+1, n in Z. The reason we do this is because it's impossible for us to list EVERY solution ourselves, but at least this way we have a simple formula (one that only includes simple addition and multiplication) that let's us generate ALL of the possible numbers
Okay, so let's put this in terms of the question you said before - what happens when you let the derivative equal 0? Well, the derivative of the function I gave you before is:
\right))
This function is also on the desmos page I sent to you - just click on the bubble next to it to make sure it's showing. Notice that every time that function equals 0, the sine function has a maximum or a minimum - I mean, you already know this, you were using this information yourself TO FIND the maximums and minimums! Well, you'll notice that this derivative function has a zero (intercepts the x-axis, solves the equation f'(x)=0 - whatever terminology you wish to put here) every time x is an integer. That is, this time, the solution is:
This is exactly half the period, because there are two solutions to the equation in the period. This isn't going to always be the case - and I'll discuss that a bit more below. For now, let's focus on the questions you're actually asking. So, what would happen if you were to put this into your CAS? You'll likely get the following answer:
When the CAS is doing this, it's telling you to do EXACTLY what we did before - to keep putting in different integers so you can list ALL of the possible answers. Why does it do this? For the exact same reason I did - it can't list every number, because there's an INFINITE number of them!
So, why don't you need to do this if the question states a domain? Well, let's work with a new desmos graph:
I'm using the same graphs this time, but instead the domain is between 0 and 6. The derivative function is listed here again if you want to see it.Okay, so now if we want to know where all the maximum are, if we look at the graph, they're when x=1, when x=3, when x=5... But then there's no more! What's the difference this time? Well, this time, the graph ENDS at x=0 and x=6 - we know that if it WERE to keep going, then the next two maxima would be at x=-1, and x=7, but the graph stops before we get to there. So this time, because the number of solutions is in a small countable list, it's easier to just list the numbers, and say that the solution is:
Because we CAN do that, and this is easier to read. The n notation stuff is a little more complicated, but it's the simplest way to list an INFINITE number of options. The easiest way to list a FINITE number of options? Is to just list them.
Finally, some side notes - in the real world, if the answer WAS simply just all the odd numbers, you could say "there is a maximum every time x is odd", and people would know what you mean. However, that number isn't always going to be made pretty - I purposefully made it pretty to make it easier to explain. If you were just solving "where are the maximums for y=sin(x)", then your general solution would be:
Which is much more confusing, and harder to explain with words. That's why we use equations, instead.
Finally, when solving equations of the form "sin(x)=0.5", you may need to use multiple lines to write your solution. The typical way to solve something of a general solution is to just find the first two solutions, and then add n*period of the function. So for our example with 0=pi*cos(pi(x-0.5)), we know that the solution is x=n - and this is because each solution is exactly half a period apart (period is 2, solutions are 1 apart). However, for sin(x)=0.5, our period is 2pi, and the first two solutions are:
And there are not pi units apart - they're much less. So in this case, our general solutions would be:
Wait, but what if we took this approach for our example graph? Well, the first two solutions are 0 and 1, and our period is 2, so the answer would be:
Which is the same as writing every even and every odd integer - which is the exact same list of numbers as every integer! In fact, if we were to write this out as a list, we'd get:
,2(0)+1=0,1\\<br />x=2(1),2(1)+1=2,3\\<br />x=2(2),2(2)+1=4,5)
And you get the idea. Even though they're written in two different ways, the answers are EXACTLY the same when you start writing out lists. Sure, x=n is more elegant, but would you lose marks if you wrote it out as x=2n,2n+1? No, you shouldn't. And anyone who says otherwise are ignoring VCAA's own advice, who say that as long as what you've done is mathematically correct, you will get the right answer.
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