Advanced Higher Maths 2011-2012 : Discussion and Help

Original post by Hype en Ecosse

I see you got the whiteboard - I'm so jelly :tongue:

I'm backing up what Jordan did there just so you know he's not talking crap, but as a personal thing, rather than writing it as if I'm dividing by the "coefficient" of the function which e is to the power of, I write it as a product of a fraction and the original exponent - I just think it looks nicer that way.

ie

\frac{1}{2}e^{2x} + c

Thought I'd chip in with my two pence.

Hell yeah. It's actually incredibly useful (especially for questions like this).

Yeah, you can do either fractional form or the way I do it - personally my handwriting doesn't accommodate for neat-looking fractions, so it's best if I leave it in that form. Plus, it's quicker to write. :tongue:

And I'm glad you can convince people I'm not talking crap. If you can teach me to do that, it'd be most appreciated.

Original post by Harley

Wow thank you so much for that :biggrin:

I'll save the image onto my computer lol. The stuff for the natural logs really helped since before it was really just guesswork when I managed to get an answer lol. I won't bother you with the curve sketching one, you've already given up loads of your time for me, I'll go have another try at it myself, I think maybe I was just panicking after the disaster that was outcome 3!

You're welcome! And good. You'd be surprised at how logical maths is when it's shown to you in a way that makes sense.

No, post the curve sketching question! Let's face it: if I had something better to do with my time, I wouldn't be here, would I? :wink:

Original post by Hype en Ecosse

I'll do the curve sketching one instead, if you like. I like doing this stuff, keeps the knowledge fresh in my head, haha. :smile:

Mine! >:c

Feel free. At least your handwriting/LaTex use won't look like a drunk spider stepped in ink and then ran across a board...

(edited 12 years ago)

Reply 301

Harley

16

Original post by JordanR

Hell yeah. It's actually incredibly useful (especially for questions like this).

Yeah, you can do either fractional form or the way I do it - personally my handwriting doesn't accommodate for neat-looking fractions, so it's best if I leave it in that form. Plus, it's quicker to write. :tongue:

And I'm glad you can convince people I'm not talking crap. If you can teach me to do that, it'd be most appreciated.

You're welcome! And good. You'd be surprised at how logical maths is when it's shown to you in a way that makes sense.

No, post the curve sketching question! Let's face it: if I had something better to do with my time, I wouldn't be here, would I? :wink:

Mine! >:c

Feel free. At least your handwriting/LaTex use won't look like a drunk spider stepped in ink and then ran across a board...

Lol ok, yous can both do it if you want :P I'll have a wee go myself and then I can compare what I get to your answers :smile:

f(x) = x^2 - x + 7 / (x - 2)

x =/= 2

Reply 302

Hype en Ecosse

20

Original post by Harley

Lol ok, yous can both do it if you want :P I'll have a wee go myself and then I can compare what I get to your answers :smile:

f(x) = x^2 - x + 7 / (x - 2)

x =/= 2

First thing you want to do is determine vertical asymptotes, and these are found at the zeroes of the denominators (at x = 2 in this case). The reason for this is that the closer you get to that 0, the further and further your result will get to infinity, you can test this by shoving in 1.9, then 1.99, then 1.999999999999, just as long as you never put in 2. And the mathematical way of representing this would be:

\displaystyle\lim_{x \to 2} f(x) = \pm \infty

Although, sometimes they're sneaky and there's a solution for your denominator that WON'T make a vertical asymptote, and you'll know it won't make a vertical asymptote if you put it into your function and you get zero over zero, if you get any other number over zero, it's definitely a VA!

So for this, we have to consider what happens when we approach the asymptote from both sides, the positive side, and negative side. Do this whatever way you've been taught, the way we do it is buy shoving in a number that is slightly smaller than your zero, and then a number that is slightly bigger than your zero, if the overall function is positive or negative, this will tell you where abouts your function approaches this asymptote. Easy!

at x \rightarrow 2^- (x < 2), y \rightarrow \frac{+a}{-a} \rightarrow -\infty

, where a is just any random number. It doesn't matter what the number is, just whether it's positive or negative.
You'd repeat this for approaching 2 from a positive direction, i.e, putting in a number a little bigger than 2, and you'll find that y approaches positive infinity this time!

So next you have to determine whether the graph has horizontal or oblique asymptotes and I have a little check list which I go through:
-Compare the degrees of the numerator and denominator:
- If the degrees are the same (i.e, the highest power is squared, over highest power is squared), then you have a horizontal asymptote at:

y = \frac{numerator's \ leading \ coefficient}{denominator's \ leading \ coefficient}

- If the denominator's degree is greater than the numerator's, you have a HA at y = 0

- If the numerator's degree is greater, by one, then you have an oblique asymptote that you need to find for long division.

Unfortunately in this case, we've got the last one, an oblique asymptote, and those always take the longest to find! So we do long division! If you can't do long division of polynomials look through these.

So you should end up with something like:

f(x) = x + 1 + \frac{9}{x-2}

And from this, y = x + 1 is our oblique asymptote!
Now we have to figure out how our curves approach this asymptote!

This is the part that I don't understand why I do, my teacher just told us to do it...divide the terms in the fraction by the highest power of x, this turns the function into:

f(x) = (x + 1) + \frac{\frac{9}{x}}{1 -\frac{2}{x}}

Now we do what we did for the vertical asymptotes, but instead, we shove in x = +infinity, and x = -infinity, and see what it comes out as approaching y at, and we get.

x \rightarrow +\infty , y \rightarrow (x + 1)^+

x \rightarrow -\infty , y \rightarrow (x + 1)^-

(What the signs above the x + 1 mean is, the plus is approaching from positive, minus is approaching from negative.

Now, if we take all of this, and finally we can move onto the easy stuff! Differentiate the original function using the quotient rule so you can gain stationary points, just like you did at higher. Do all this, and you'll find

f'(x) = \frac{x^2 - 4x - 5}{(x - 2)^2}

Set this equal to zero, and we can make the bottom disappear, leaving us with a quadratic to solve: (x - 5)(x + 1) = 0.
Shove these points of x into the original function to get y values and we find:
x = 5: y = 9
x = -1: y = - 3

Those are our stationary points, now to find their nature. You can do this by finding the second derivative, or by drawing a nature table. I like the second derivative way.
If f''(x) < 0, maximum turning point
If f''(x) > 0, minimum turning point. By shoving in our values of x, we get:

(5,9) is a minimum turning point
(-1, -3) is a maximum turning point.

This is you with ALL the information you need to draw your curve. Just sketch on the asymptotes clearly, and what I like to do is check for a y-intercept by putting 0 into x so that f(0) = y-intercept, this curve intercepts the y axis at

-\frac{7}{2}

If you've drawn this all onto your graph correctly it should look something like this:

(this excludes the asymptotes, but you should ALWAYS draw them on :smile:

)

(edited 12 years ago)

Reply 303

Harley

16

[QUOTE="Ecosse;34718049" Hype="Hype" en="en"]First thing you want to do is determine vertical asymptotes, and these are found at the zeroes of the denominators (at x = 2 in this case). The reason for this is that the closer you get to that 0, the further and further your result will get to infinity, you can test this by shoving in 1.9, then 1.99, then 1.999999999999, just as long as you never put in 2. And the mathematical way of representing this would be:

\displaystyle\lim_{x \to 2} f(x) = \pm \infty

Although, sometimes they're sneaky and there's a solution for your denominator that WON'T make a vertical asymptote, and you'll know it won't make a vertical asymptote if you put it into your function and you get zero over zero, if you get any other number over zero, it's definitely a VA!

So for this, we have to consider what happens when we approach the asymptote from both sides, the positive side, and negative side. Do this whatever way you've been taught, the way we do it is buy shoving in a number that is slightly smaller than your zero, and then a number that is slightly bigger than your zero, if the overall function is positive or negative, this will tell you where abouts your function approaches this asymptote. Easy!

at x \rightarrow 2^- (x < 2), y \rightarrow \frac{+a}{-a} \rightarrow -\infty

, where a is just any random number. It doesn't matter what the number is, just whether it's positive or negative.
You'd repeat this for approaching 2 from a positive direction, i.e, putting in a number a little bigger than 2, and you'll find that y approaches positive infinity this time!

So next you have to determine whether the graph has horizontal or vertical asymptotes and I have a little check list which I go through:
-Compare the degrees of the numerator and denominator:
- If the degrees are the same (i.e, the highest power is squared, over highest power is squared), then you have a horizontal asymptote at:

y = \frac{numerator's \ leading \ coefficient}{denominator's \ leading \ coefficient}

- If the denominator's degree is greater than the numerator's, you have a HA at y = 0

- If the numerator's degree is greater, by one, then you have an oblique asymptote that you need to find for long division.

Unfortunately in this case, we've got the last one, an oblique asymptote, and those always take the longest to find! So if we do long division! If you can't do long division of polynomials look through these.

So you should end up with something like:

f(x) = x + 1 + \frac{9}{x-2}

And from this, y = x + 1 is our oblique asymptote!
Now we have to figure out how our curves approach this asymptote!

This is the part that I don't understand why I do, my teacher just told us to do it...divide the terms in the fraction by the highest power of x, this turns the function into:

f(x) = (x + 1) + \frac{\frac{9}{x}}{1 -\frac{2}{x}}

Now we do what we did for the vertical asymptotes, but instead, we shove in x = +infinity, and x = -infinity, and see what it comes out as approaching y at, and we get.

x \rightarrow +\infty , y \rightarrow (x + 1)^+

x \rightarrow -\infty , y \rightarrow (x + 1)^-

(What the signs above the x + 1 mean is, the plus is approaching from positive, minus is approaching from negative.

Now, if we take all of this, and finally we can move onto the easy stuff! Differentiate the original function using the quotient rule so you can gain stationary points, just like you did at higher. Do all this, and you'll find

f'(x) = \frac{x^2 - 4x - 5}{(x - 2)^2}

Set this equal to zero, and we can make the bottom disappear, leaving us with a quadratic to solve: (x - 5)(x + 1) = 0.
Shove these points of x into the original function to get y values and we find:
x = 5: y = 9
x = -1: y = - 3

Those are our stationary points, now to find their nature. You can do this by finding the second derivative, or by drawing a nature table. I like the second derivative way.
If f''(x) < 0, maximum turning point
If f''(x) > 0, minimum turning point. By shoving in our values of x, we get:

(5,9) is a minimum turning point
(-1, -3) is a maximum turning point.

This is you with ALL the information you need to draw your curve. Just sketch on the asymptotes clearly, and what I like to do is check for a y-intercept by putting 0 into x so that f(0) = y-intercept, this curve intercepts the y axis at

Unparseable latex formula:

-\frac{7}{2}[br][br]If you've drawn this all onto your graph correctly it should look something like this: [br][img]http://www3.wolframalpha.com/Calculate/MSP/MSP223919hia59i10i4ab4600000gge92e9h9d7a696?MSPStoreType=image/gif&s=33&w=306&h=126&cdf=RangeControl[/img] (this excludes the asymptotes, but you should ALWAYS draw them on [s]smile[/s] )

Yay I think I did it right then! I did it just there and my graph looks the same as yours and I did the same calculations so I think i'll actually get some marks this time! I'll just need to hope that I can get it all worked out and drawn in time lol.

Thanks so much for all your help :biggrin:

, that must have been a nightmare to type!

Reply 304

ukdragon37

Original post by Hype en Ecosse

Unfortunately in this case, we've got the last one, an oblique asymptote, and those always take the longest to find! So if we do long division! If you can't do long division of polynomials look through these.

So you should end up with something like:

f(x) = x + 1 + \frac{9}{x-2}

And from this, y = x + 1 is our oblique asymptote!
Now we have to figure out how our curves approach this asymptote!
...

Now, if we take all of this, and finally we can move onto the easy stuff! Differentiate the original function using the quotient rule so you can gain stationary points, just like you did at higher. Do all this, and you'll find

f'(x) = \frac{x^2 - 4x - 5}{(x - 2)^2}

...

Good work. :smile:

Just one comment. Once you've performed polynomial long division it's usually easier to differentiate the result than to differentiate the original function, so in this case you'll get:

f(x) = x + 1 + \dfrac{9}{x-2}

f'\left( x \right) = 1 - \dfrac{9}{\left(x-2\right)^2} = 0

\left(x-2\right)^2 = 9

Also In the part where you are dividing by the highest powers of x, it guarantees to result in terms dividing by x which would tend to 0 as x tends to infinity.

Reply 305

It seems he's worth all the Hype.

Original post by ukdragon37

Good work. :smile:

Just one comment. Once you've performed polynomial long division it's usually easier to differentiate the result than to differentiate the original function, so in this case you'll get:

f(x) = x + 1 + \dfrac{9}{x-2}

f'\left( x \right) = 1 - \dfrac{9}{\left(x-2\right)^2} = 0

\left(x-2\right)^2 = 9

Also In the part where you are dividing by the highest powers of x, it guarantees to result in terms dividing by x which would tend to 0 as x tends to infinity.

Yep, was gonna say this. Saves a lot of time.

Reply 306

Hype en Ecosse

20

Original post by ukdragon37

Good work. :smile:

Just one comment. Once you've performed polynomial long division it's usually easier to differentiate the result than to differentiate the original function, so in this case you'll get:

f(x) = x + 1 + \dfrac{9}{x-2}

f'\left( x \right) = 1 - \dfrac{9}{\left(x-2\right)^2} = 0

\left(x-2\right)^2 = 9

Also In the part where you are dividing by the highest powers of x, it guarantees to result in terms dividing by x which would tend to 0 as x tends to infinity.

Thanks a lot for the input :smile:

Especially for the last bit, I've asked my teacher on 3 separate occasions and looked through multiple textbooks, but none of them gave the answer. Just waffled really. :tongue:

And I know it's easier to differentiate the divided function, but I like to minimalise the amount of errors that could creep in. For example, if I did the long division wrong - I could at least still find the right stationary points by differentiating the original function. :smile:

Reply 307

Complex numbers are pretty cool. Anyone else started them yet?

NAB +Extension Test for Unit 1 next week for me. Kind of looking forward to doing that compared to the MAT tomorrow :s-smilie:

Reply 308

Original post by laughylolly

Complex numbers are pretty cool. Anyone else started them yet?

NAB +Extension Test for Unit 1 next week for me. Kind of looking forward to doing that compared to the MAT tomorrow :s-smilie:

i have

Reply 309

Original post by JordanR

i have

Hahaha.

Reply 310

LonelySoul193

1

Got my Unit 1 NAB result back with 100% :biggrin:

Very pleased. Though... extended NAB next week :redface:

Reply 311

Hype en Ecosse

20

Original post by laughylolly

Complex numbers are pretty cool. Anyone else started them yet?

NAB +Extension Test for Unit 1 next week for me. Kind of looking forward to doing that compared to the MAT tomorrow :s-smilie:

Ugh, no. I hate them, find them so boring! Just finished them today, and the only bit that interested me was questions on De Moivre's that required some problem solving. We were told to work through unit 2 backwards, so we've done Elementary Number Theory, Complex Numbers, and Series and Sequences. I can't wait to get back to the calculus stuff now. :tongue:

Reply 312

Right, came across this question today and kinda have no idea how to differentiate it.

f(x)=sec^{-1}3x

would this be true?

f(x)=\frac{1}{cos^{-1}3x}

then would you use quotient + chain rule or just extended chain rule? Hints please =)

(edited 12 years ago)

Reply 313

Original post by laughylolly

Right, came across this question today and kinda have no idea how to differentiate it.

f(x)=sec^{-1}3x

(This being the inverse of sec not 1/sec. )

would this be true?

f(x)=\frac{1}{cos^{-1}3x}

then would you use quotient + chain rule or just extended chain rule? Hints please =)

Well it's not going to be negative (obviously!).

If you differentiate 1/arccos(x) then you're going to get something quite different from the derivative of arcsecant(x). I can't think how to hint at this, so I'll just say it. :tongue:

Where you'd have sqrt(1-x^2) on the denominator for arccos(x), for arcsecant(x) you'd have sqrt(1-1/x^2) on the denominator.

Couldn't be bothered with LaTex so I hope that makes sense. Just in and I'm tired. :tongue:

Edit: remember that arcsecantx is just arccos(1/x). That probably helps more than what I said before.

sec^{-1}(x)= cos^{-1}(\frac{1}{x})

(edited 12 years ago)

Reply 314

Original post by JordanR

Edit: remember that arcsecantx is just arccos(1/x). That probably helps more than what I said before.

sec^{-1}(x)= cos^{-1}(\frac{1}{x})

ah okay, that I can do. Thank you!

Okay I differentiated and got

f'(x)=\frac{1}{3x^{2}\sqrt{1-\frac{1}{9x^{2}}}}

but the answer in the back of the book says it's

f'(x)=\frac{1}{x\sqrt{9x^{2}-1}}

is there a way of rearranging my answer to get the answer in the back of the book or did I differentiate wrong?

(edited 12 years ago)

Reply 315

Original post by laughylolly

ah okay, that I can do. Thank you!

No problem. And obviously

sec^{-1}(3x)=cos^{-1}(\frac{1}{3x})

And like I said before, your result's gonna be positive rather than negative, whereas if you were to take the derivative of

csc^{-1}(3x)

you'd get a negative result, which is obviously the opposite signs compared to the derivatives for inverse sine and cosine.

God I'm a nerd. :tongue:

Reply 316

Original post by JordanR

No problem. And obviously

sec^{-1}(3x)=cos^{-1}(\frac{1}{3x})

And like I said before, your result's gonna be positive rather than negative, whereas if you were to take the derivative of

csc^{-1}(3x)

you'd get a negative result, which is obviously the opposite signs compared to the derivatives for inverse sine and cosine.

God I'm a nerd. :tongue:

Ah wait never mind now, I got it. I realized what I have to do with the rearranging too. I don't like rearranging with the square root there. Makes things awkward.

Alright a challenge for anyone who wants to do it. I found this kinda tricky although you guys might find it straight forward.

Differentiate this and simplify as far as you can.

f(x)=sin^{-1}(tanx)

Spoiler

(edited 12 years ago)

Reply 317

Original post by laughylolly

Ah wait never mind now, I got it. I realized what I have to do with the rearranging too. I don't like rearranging with the square root there. Makes things awkward.

Aye, it's a bit tricky. I don't think you're likely to see anything like that in the exam though.

If you wanted another idea, try this:

y=sec^{-1}(3x)

Then

x=\frac{sec(y)}{3}

You could try differentiating implicitly then, but I don't know how much easier it'd be (if at all). Just an idea.

Original post by laughylolly

Ah wait never mind now, I got it. I realized what I have to do with the rearranging too. I don't like rearranging with the square root there. Makes things awkward.

Alright a challenge for anyone who wants to do it. I found this kinda tricky although you guys might find it straight forward.

Differentiate this and simplify as far as you can.

f(x)=sin^{-1}(tanx)

Spoiler

Will have a quick bash at that, but I have to wonder why you didn't write it as

\frac{sec(x)}{\sqrt{cos2x}}

(edited 12 years ago)

Reply 318

Original post by JordanR

Aye, it's a bit tricky. I don't think you're likely to see anything like that in the exam though.

If you wanted another idea, try this:

y=sec^{-1}(3x)

Then

x=\frac{sec(y)}{3}

You could try differentiating implicitly then, but I don't know how much easier it'd be (if at all). Just an idea.

Oh yeah you could do that... this exercise was all about using like the standard derivatives so "/

Reply 319