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March 29, 2024, 08:43:47 pm

Author Topic: VCE Methods Question Thread!  (Read 4803338 times)  Share 

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Jinju-san

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Re: VCE Methods Question Thread!
« Reply #19065 on: January 27, 2021, 10:29:51 pm »
+8
Hi, i just wanted to clarify some things I'm not sure about. Is it that e^x cannot equal zero or must be greater than zero or both? Also, would anybody be able to explain why?
Thanks!

Hi jasmine24!

So from what I know, the number e (Euler’s number which is approximately equal to 2.71828) cannot be equal to zero and must be greater than zero. This rule applies to any exponential expressions and functions. Whenever you substitute any given value of x, positive or negative, your answer must always be positive.

Let’s think about it this way:
Let  y = e^x,
Substitute x= -1:
y =  e^-1 = 1/(e^1). Since e^1 = 2.71828, 1/e > 0 (it must be a positive value).

Substitute x = 0:
y = e^0 = 1 > 0 (again this has given us a positive value).

Substitute x = 1:
y = e^ 1 = 2.71828 > 0 (once again, this has produced a positive value).

As you can see, any values of x that we substitute into y = e^x (whether positive or negative) always gives us a positive value, but never zero. Therefore, we can conclude that e^x can never be equal to zero AND must be a positive value.

Hope this helps!  ;D

p0kem0n21

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Re: VCE Methods Question Thread!
« Reply #19066 on: January 27, 2021, 10:43:38 pm »
+8
Hi, i just wanted to clarify some things I'm not sure about. Is it that e^x cannot equal zero or must be greater than zero or both? Also, would anybody be able to explain why?
Thanks!

The value of ex, for any real number x, will always be greater than zero (and cannot be equal to zero). One way you can confirm this is by simply sketching a graph of y=ex, where you will notice that the graph is always above the x-axis (y=0), but approaches y=0 (i.e. gets closer and closer to 0, but never really is zero).

But how do we explain this conceptually? Well, it's first important to take note that e is really just a positive number, 2.72 (to 2 d.p.). The second thing to realize is that we are just applying index laws to this arbitrary positive number, where the power applied is x. I am going to apply 4 different cases to hopefully help you understand why the above explanation is the case. The first and simplest case is when x>=1. In this case, the value of ex will either be e (when x=1, since a number to the power of 1 is the number itself) OR it will be greater than e. You can see this for yourself by plugging in values of x. e2 is approximately 7.39, e1.5 is 4.48, and you could go on and on just plugging in values of x which are greater than 1 for you to see that they will always be greater than e, and thus greater than zero.

The second case is 0 < x < 1. In this case, you will notice that the value of ex will actually be lower than the value of e since we are applying a power which is less than one. Once again, you can try plugging in random values of x between 0 and 1 to notice this. However, at x=0 (our third case), the value of ex is 1 since virtually all numbers to the power of 0 are 0 (we don't talk about 00 lol). Thus, returning to our second case, we can see that the value of ex will be between 1 and e when x is between 0 and 1. In other words, for both the second and third cases, ex is greater than zero.

Our fourth and final case is when x<0, when it is a negative number. Basically, as x gets lower and lower, the value of ex will become closer and closer to zero. We can remember this due to our index law for negative exponents: a-b = 1/ab.
Now in this case, our negative number is x. For example, let x=-1. Then, e-1 = 1/e. Keep in mind that this is a positive number divided by a positive number, so it cannot be a negative number. This applies for any negative value of x. Furthermore, note that the only possible way for this to equate to zero is if the numerator is equal to zero, which is not possible here. That explains why it will still always be greater than zero.
But why does it approach zero as x becomes greater? Essentially, it's because you're dividing 1 by a larger number as x decreases further. To illustrate this, compare x=-1 (e-1, or 1/e1), x=-2 (e-2, or 1/e2), and x=-3 (e-3, or 1/e3). You can see that as x becomes more negative, we are dividing by a larger number which makes the value really close to zero, but never zero.

Hopefully this all makes sense  :)

Edit: beat by Jinju-san, but I'll still post this in case it's any help haha

Samueliscool223

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Re: VCE Methods Question Thread!
« Reply #19067 on: February 06, 2021, 03:52:04 pm »
0
why does the domain f o g equal domain of g?

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #19068 on: February 06, 2021, 06:04:23 pm »
+4
why does the domain f o g equal domain of g?

Remember that \(f\circ g(x)=f(g(x))\). Another way of thinking of this is the idea of a set of machines. Let's say that g(x) is a machine that takes a number x and turns it into u (that is, \(u=g(x)\)). Then, f(u) is a machine that takes a number u and turns it into y (that is, \(y=f(u)\)). If a number is "outside of the domain" of either of these numbers, then let's pretend that the number doesn't fit into the machine. For example, if you had a machine that accepted apples and turned them into oranges, it probably won't be big enough to fit a watermelon, or bananas might be a little too long for it to take them in. Same idea with numbers needing to be in the domain for your function to compute them - if g(x) is only defined for [2,3], then 4 is too big for it, or maybe 0 is too long to fit in our machine. Whatever the reason, the number just won't for into the machine that is our function.

So, what does this have to do with fog(x)? Well, let's say a number is outside the domain of g. If the number doesn't fit into g, then g can't provide an output u. If g can't make a u, then there is no u to go into the second machine, f(u). If u doesn't go into f(u), then you can't get a y at the end. As a result, the domain of fog(x) ends up being the same as the domain of g(x).

Samueliscool223

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Re: VCE Methods Question Thread!
« Reply #19069 on: February 07, 2021, 09:47:00 am »
0
Remember that \(f\circ g(x)=f(g(x))\). Another way of thinking of this is the idea of a set of machines. Let's say that g(x) is a machine that takes a number x and turns it into u (that is, \(u=g(x)\)). Then, f(u) is a machine that takes a number u and turns it into y (that is, \(y=f(u)\)). If a number is "outside of the domain" of either of these numbers, then let's pretend that the number doesn't fit into the machine. For example, if you had a machine that accepted apples and turned them into oranges, it probably won't be big enough to fit a watermelon, or bananas might be a little too long for it to take them in. Same idea with numbers needing to be in the domain for your function to compute them - if g(x) is only defined for [2,3], then 4 is too big for it, or maybe 0 is too long to fit in our machine. Whatever the reason, the number just won't for into the machine that is our function.

So, what does this have to do with fog(x)? Well, let's say a number is outside the domain of g. If the number doesn't fit into g, then g can't provide an output u. If g can't make a u, then there is no u to go into the second machine, f(u). If u doesn't go into f(u), then you can't get a y at the end. As a result, the domain of fog(x) ends up being the same as the domain of g(x).

thanks for your reply but i still dont quite get it...
wouldnt it make sense if the range of g(x) was the domain of f o g? because doesnt the output values of g serve as the input values for fog and hence would be the domain?
e.g. say, function minimum is g(3) = 5, function maximum is g(7) = 15, then the domain of fog  would be [5, 15], and not [3, 7], no??
« Last Edit: February 07, 2021, 10:29:05 am by Samueliscool223 »

makram

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Re: VCE Methods Question Thread!
« Reply #19070 on: February 07, 2021, 12:30:53 pm »
+4
Remember, we need to actually plug in our original x values into g(x), and then the output is used as an input to f(x). Therefore, [3,7] is the domain of the composite function, as you can only input values between 3 and 7 first into g(x) and then the output of that [5,15] can be inputted into f(x).

Suppose g(x)=√x and f(x)=2x^3+x+2, .: f(g(x))= 2(√x)^3+(√x)+2
Here, we can only input x values that are within the domain of √x into the entire composite function, as every x value in f(x) becomes √x.
(Working with real numbers) Lets attempt to plug in x=-1 into f(g(x)):   f(g(x))=2(√-1)^3+√-1+2    Hence,  x=-1 is undefined for the entire composite function as it is also undefined for g(x)=√x

Therefore, if you cannot plug in x values into g(x) you cannot plug it into f(g(x)) and thus the domain of g(x) is also the domain of f(g(x))

jasmine24

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Re: VCE Methods Question Thread!
« Reply #19071 on: February 07, 2021, 06:55:33 pm »
0
How do you determine the steepest point on a function? I tried using optimisation but that gave a gradient of 0 and idk how else to find it

p0kem0n21

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Re: VCE Methods Question Thread!
« Reply #19072 on: February 07, 2021, 07:41:56 pm »
+7
How do you determine the steepest point on a function? I tried using optimisation but that gave a gradient of 0 and idk how else to find it

I assume that when you say steepest point, you mean the point with the highest gradient. Let's use the arbitrary function f(x). We already know that we can find the gradient at a particular point using f'(x), and that when f'(z) = 0, there is a stationary point at z (z is just an arbitrary variable I subbed in). But what about finding the point of highest gradient?

There's two main things we need to consider. The first is the derivative of f'(x), or f''(x). This function shows us the change in gradient. Similarly to how f'(x) = 0 shows a minimum/maximum turning point (sometimes a point of inflection), f''(x)=0 shows us the minimum/maximum gradient. So one thing you will have to do is find f''(x), solve f''(x) = 0 for x, and see whether this value of x corresponds to a 'minimum' or 'maximum' for gradient (the method for doing this is similar to how you determine if a turning point is maximum or minimum).

The second thing to consider is the domain of the function. For example, consider f(x) over a restricted domain. You find x such that f''(x) = 0, but realise that this is a "minimum" for the gradient as opposed to the "steepest" point/maximum of gradient. Or maybe you can't find a value for x so that f''(x) = 0 (within the domain of the function). Then what? In these cases, chances are you have to look at the gradients of the endpoints of the function (that is to say, at the ends of the function). Now, you can't really have a gradient at the endpoints of a function for Methods (which is why you are probably unlikely to reach this stage if you have considered all of the above), but if we could, then these are the points where you would expect to find the greatest/least gradient.

Hope this helps  :)

jasmine24

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Re: VCE Methods Question Thread!
« Reply #19073 on: February 08, 2021, 08:12:03 am »
0
I assume that when you say steepest point, you mean the point with the highest gradient. Let's use the arbitrary function f(x). We already know that we can find the gradient at a particular point using f'(x), and that when f'(z) = 0, there is a stationary point at z (z is just an arbitrary variable I subbed in). But what about finding the point of highest gradient?

There's two main things we need to consider. The first is the derivative of f'(x), or f''(x). This function shows us the change in gradient. Similarly to how f'(x) = 0 shows a minimum/maximum turning point (sometimes a point of inflection), f''(x)=0 shows us the minimum/maximum gradient. So one thing you will have to do is find f''(x), solve f''(x) = 0 for x, and see whether this value of x corresponds to a 'minimum' or 'maximum' for gradient (the method for doing this is similar to how you determine if a turning point is maximum or minimum).

The second thing to consider is the domain of the function. For example, consider f(x) over a restricted domain. You find x such that f''(x) = 0, but realise that this is a "minimum" for the gradient as opposed to the "steepest" point/maximum of gradient. Or maybe you can't find a value for x so that f''(x) = 0 (within the domain of the function). Then what? In these cases, chances are you have to look at the gradients of the endpoints of the function (that is to say, at the ends of the function). Now, you can't really have a gradient at the endpoints of a function for Methods (which is why you are probably unlikely to reach this stage if you have considered all of the above), but if we could, then these are the points where you would expect to find the greatest/least gradient.

Hope this helps  :)
Thank u so much! this helps alot  :)

Writing Overload

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Re: VCE Methods Question Thread!
« Reply #19074 on: February 11, 2021, 03:05:12 pm »
0
Hello AN,

Can someone help me with this question?

Thanks!


fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19075 on: February 11, 2021, 04:10:01 pm »
+7
Usually with questions like this you want to be manipulating the expression in such a way that the value of the expression doesn't change, but the form of the expression is 'nicer'/ more recognisable. Examples include multiplying and dividing by the same number, adding and subtracting the same number from the denominator, etc.

Here, we'll use the latter:


Hope this helps :)
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Writing Overload

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Re: VCE Methods Question Thread!
« Reply #19076 on: February 11, 2021, 04:39:19 pm »
0
Usually with questions like this you want to be manipulating the expression in such a way that the value of the expression doesn't change, but the form of the expression is 'nicer'/ more recognisable. Examples include multiplying and dividing by the same number, adding and subtracting the same number from the denominator, etc.

Here, we'll use the latter:


Hope this helps :)

Ahh yep got it thanks! Is this like one of those skills that you grow from practicing?

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19077 on: February 11, 2021, 05:01:07 pm »
+7
Yeah, definitely. It helps to get a lot of practice in with techniques like this because it's hardly ever explicitly mentioned or even implied - much like you're expected to go from A to B when proving something (you're really only given the structure of the proof (ie. induction, contradiction, etc)). If you get enough practice in you'll recognise what to do and when pretty quickly.

imo there's also a couple of ways to really go about this:
- tinker with the expression until you find a more recognisable form
- think about what you're working towards then manipulate the expression in that direction

Personally, I prefer the second method, but it doesn't always work (especially if you can't find a clear goal or there isn't one at all). Try to get more practice in, and if you need help let us know :D
« Last Edit: February 11, 2021, 05:03:01 pm by fun_jirachi »
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Bluebird

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Re: VCE Methods Question Thread!
« Reply #19078 on: February 15, 2021, 07:08:37 pm »
0
Hey AN,

How do you figure out which way to write your signs when solving for quadratic inequalities? Like, how do you figure out whether the region required is inside the parabola or outside the parabola? (I hope I phrased this correctly)

Sine

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Re: VCE Methods Question Thread!
« Reply #19079 on: February 15, 2021, 07:29:51 pm »
+6
Hey AN,

How do you figure out which way to write your signs when solving for quadratic inequalities? Like, how do you figure out whether the region required is inside the parabola or outside the parabola? (I hope I phrased this correctly)
Draw a diagram, I think that is the best way of solving non-linear inequalities since it is so easy to mess up when just manipulating the inequality by itself.