First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

He-Mann · Apr 20, 2017

Re: MATH1131 help thread

[R] Requires IVT (not hard).

[H] Write the polynomial as: p₃(x) = p₂(x) + x³/3!. Then use previous part and calculus (monotonicity) to argue that p₃ has exactly one root.

leehuan · Apr 20, 2017

Re: MATH1131 help thread

matchalolz said:
View attachment 33883

Do I need to find some sort of interval to use MVT or IVT?

Yes, you do need to find an interval to use the IVT for [R]. But since the question just specifies that at least one root must exist, nobody is going to care what interval you pick. You just have to pick any interval that works.

[H] is different because you're trying to prove that no root exists. Hint: Think 2U, because you're given a quadratic.
You can then use what He-Mann suggested.

matchalolz · Apr 20, 2017

Re: MATH1131 help thread

He-Mann said:
[R] Requires IVT (not hard).

[H] Write the polynomial as: p₃(x) = p₂(x) + x³/3!. Then use previous part and calculus (monotonicity) to argue that p₃ has exactly one root.

Ok so I got the first part with IVT as you suggested,

with the second part, I concluded that p2(x) has only one stationery point (-1,1/2) and its concave up, hence it can't ever touch the x axis. So can you just say because p2(x) is always positive, p3(x) can only have one root?

matchalolz · Apr 20, 2017

Re: MATH1131 help thread

leehuan said:
Yes, you do need to find an interval to use the IVT for [R]. But since the question just specifies that at least one root must exist, nobody is going to care what interval you pick. You just have to pick any interval that works.

[H] is different because you're trying to prove that no root exists. Hint: Think 2U, because you're given a quadratic.
You can then use what He-Mann suggested.

How would you know whether to use IVT or MVT? I think I'm kind of confused over that

InteGrand · Apr 20, 2017

Re: MATH1131 help thread

matchalolz said:
Ok so I got the first part with IVT as you suggested,

with the second part, I concluded that p2(x) has only one stationery point (-1,-1/2) and its concave up, hence it can't ever touch the x axis. So can you just say because p2(x) is always positive, p3(x) can only have one root?

$\noindent You can say that since $p_{3}'(x) = p_{2}(x) > 0$ for all $x\in \mathbb{R}$, the function $p_{3}$ is increasing, so cannot have two real roots. (Alternatively, if it had two real roots, Rolle's Theorem would imply the existence of a real root of $p_{2}$, but we know such a root does not exist.)$

leehuan · Apr 20, 2017

Re: MATH1131 help thread

matchalolz said:
How would you know whether to use IVT or MVT? I think I'm kind of confused over that

For starters
The IVT is related to f
The MVT is related to f'
Whilst obvious, this saved me a bit of thinking a few times.

Perhaps more clearly, very rarely do you ever use the MVT to show the existence of the root. The IVT wants you to examine the actual function values, so it's the more suitable one. The MVT is considering the gradient.

matchalolz · Apr 20, 2017

Re: MATH1131 help thread

leehuan said:
For starters
The IVT is related to f
The MVT is related to f'
Whilst obvious, this saved me a bit of thinking a few times.

Perhaps more clearly, very rarely do you ever use the MVT to show the existence of the root. The IVT wants you to examine the actual function values, so it's the more suitable one. The MVT is considering the gradient.

I'll keep that in mind

It's probably going to take some time before this fully sinks into my mind

matchalolz · Apr 25, 2017

Re: MATH1131 help thread

Screen Shot 2017-04-25 at 4.35.55 pm.png

i can't use l'hopital's rule for this can I? How am I meant to know what sin(1/x) even looks like in the first place?

He-Mann · Apr 25, 2017

Re: MATH1131 help thread

matchalolz said:
View attachment 33904

i can't use l'hopital's rule for this can I? How am I meant to know what sin(1/x) even looks like in the first place?

Let u = 1/x and see what happens when you find the limit.

InteGrand · Apr 25, 2017

Re: MATH1131 help thread

matchalolz said:
View attachment 33904

i can't use l'hopital's rule for this can I? How am I meant to know what sin(1/x) even looks like in the first place?

$\noindent Rewrite the function as $\frac{\sin \frac{1}{x}}{\frac{1}{x}}$. This tends to $1$ as $x \to \infty$, using the limit result $\lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$. Hence the answer is $1$.$

matchalolz · Apr 25, 2017

Re: MATH1131 help thread

InteGrand said:
$\noindent Rewrite the function as $\frac{\sin \frac{1}{x}}{\frac{1}{x}}$. This tends to $1$ as $x \to \infty$, using the limit result $\lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$. Hence the answer is $1$.$

Wait I'm kind of confused, after you rewrite the function, do you use l'hopital's rule on it? or do you just know that it tends to one (?)

matchalolz · Apr 25, 2017

Re: MATH1131 help thread

Wait never mind, I kind of get it in my head but I don't know how to write the working out formally

InteGrand · Apr 25, 2017

Re: MATH1131 help thread

matchalolz said:
Wait I'm kind of confused, after you rewrite the function, do you use l'hopital's rule on it? or do you just know that it tends to one (?)

matchalolz said:
Wait never mind, I kind of get it in my head but I don't know how to write the working out formally

$\noindent The result $\lim_{\theta \to 0}\frac{\sin \theta}{\theta}$ is a standard result. Basically use the substitution $\theta = \frac{1}{x}$, so $\theta \to 0^{+}$ as $x \to \infty$ and the limit becomes $\lim_{\theta \to 0^{+}}\frac{\sin \theta}{\theta} = 1$.$

matchalolz · Apr 25, 2017

Re: MATH1131 help thread

Thanks guys :')

lmao I'm so dumb

matchalolz · Apr 29, 2017

Re: MATH1131 help thread

Screen Shot 2017-04-29 at 4.40.15 pm.png

How do I do this?

InteGrand · Apr 29, 2017

Re: MATH1131 help thread

matchalolz said:
View attachment 33905

How do I do this?

$\noindent What is the question? Sketch that curve? Simplify the expression? Anyway, just note that $\sin^{-1}(\sin x) = \pi - x$ for all $\frac{\pi}{2}\leq x \leq \frac{3\pi}{2}$. To see this, you should try and make sure you understand the sine function's periodic nature / symmetry, and the definition of $\sin^{-1}(\cdot)$.$

matchalolz · Apr 29, 2017

Re: MATH1131 help thread

InteGrand said:
$\noindent What is the question? Sketch that curve? Simplify the expression? Anyway, just note that $\sin^{-1}(\sin x) = \pi - x$ for all $\frac{\pi}{2}\leq x \leq \frac{3\pi}{2}$. To see this, you should try and make sure you understand the sine function's periodic nature / symmetry, and the definition of $\sin^{-1}(\cdot)$.$

The question was to simplify the expression.

are you subtracting x from 180 degrees because the range of the the sine inverse function is restricted to between -90 and 90 degrees?

InteGrand · Apr 29, 2017

Re: MATH1131 help thread

matchalolz said:
The question was to simplify the expression.

are you subtracting x from 180 degrees because the range of the the sine inverse function is restricted to between -90 and 90 degrees?

$\noindent Pretty much (and because this gives the correct value). As an exercise, you may wish to try the following: simplify $\cos^{-1}(\cos x)$ for $\pi < x < \frac{3\pi}{2}$.$

matchalolz · Apr 29, 2017

Re: MATH1131 help thread

InteGrand said:
$\noindent Pretty much (and because this gives the correct value). As an exercise, you may wish to try the following: simplify $\cos^{-1}(\cos x)$ for $\pi < x < \frac{3\pi}{2}$.$

i dont think I really understand this

would guess -(Pi-x)??

InteGrand · Apr 29, 2017

Re: MATH1131 help thread

matchalolz said:
i dont think I really understand this

would guess -(Pi-x)??

$\noindent That's not the right answer I'm afraid. Basically, in general, for any $u\in [-1,1]$, $\cos^{-1}u$ is the angle in the range $[0,\pi]$ whose cosine is $u$ (such an angle exists and is unique for any $u \in [-1,1]$). So $\cos^{-1}(\cos x)$ is that angle in the range $[0,\pi]$ whose cosine is $\cos x$. In other words, it's an angle ``like'' $x$ (since the cosine of $x$ is also $\cos x$), but can't be $x$ itself if $\pi < x < \frac{3\pi}{2}$, because this is not in the range $[0,\pi]$. So the answer would be the angle $\theta$ between $0$ and $\pi$ that has $\cos \theta = \cos x$. You can use, for example, general solutions of trig. equations from HSC 3U maths to find this $\theta$.$

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