leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

laters · Mar 5, 2016

leehuan said:
Only the case for (-1,0) but the whole question is again given (with a previous part)

$\\ $a) Prove that $f\left(x\right)=1+x+{x}^{2}$ is positive for all $x. \\ $b) By considering the cases $x\ge 0, \, \bf{-1<x<0}, \, $and $x\le -1\\ $prove that $1+x+{x}^{2}+{x}^{3}+{x}^{4} $ is always positive.$$

You must be doing math1151 too?? haha I got stuck on this for a bit too. I just used a GP summation

$1+x+x^2+x^3+x^4=\frac{x^5-1}{x-1}$

since we're not considering this when x=1. then the top is negative, and so is the bottom! so when you divide it it's positive.

Of course it would work for basically all the cases except x=1 where you would have to sub it in the original. it's just 5 which is positive anyway

EDIT: woops just saw it was already done by integrand, pls ignore

seanieg89 · Mar 5, 2016

And the relationship between these conditions and invertibility:

A function is f is surjective <=> it has a right inverse.

A function f is injective <=> it has a left inverse.

And a function f is bijective iff it is invertible.

Paradoxica · Mar 5, 2016

leehuan said:
Lol thanks everyone. I ended up doing this:

Let x=-b, 0<b<1
RTP:1-b+b^2-b^3+b^4

1>b implies 1-b>0
b>0 allows multiplying by b^3: b^4-b^3>0
b>0 also implies b^2>0
Sum them up.
________________

Getting confused with what I learnt. Someone please break down surjective functions for me, and injective just means one-to-one both x and y (therefore inverse function exists for the maximum domain) right?

Injective: For every value of x in the domain, there exists one unique corresponding value for y in the codomain.

Surjective: For every value of y in the codomain, there exists at least one corresponding value of x in the domain.

leehuan · Mar 5, 2016

Oh cool. Reading both of your definitions cleared it up for me much more. Thanks again!

laters said:
You must be doing math1151 too??

Lol just to clear that up, look at my sig

leehuan · Mar 5, 2016

Useless post in my own thread: Lol I feel so dumb at maths now.
___________________

Notation question:

$\\ $My lecturer has demonstrated that the notation Ran$\left(f\right) $ may be used to denote the range of a function $f$. \\ Just want to ensure that this is fine, and if there is a corresponding notation for domain?$$

Carrotsticks · Mar 5, 2016

http://i0.kym-cdn.com/photos/images/original/000/002/116/Longcat_War.jpg

leehuan · Mar 5, 2016

More stupid questions... I should just chill with the maths for tonight but.

1. How would I prove that a function is injective and surjective? Trying to prove that f(x)=x^3+3x+1 has an inverse over \mathbb R. (Without graphical approach)

All I thought about is f'(x)=3x^2+3 is positive so f(x) monotonic increasing. Is that enough?
____________

And

2. Forgot how to use dy/dx.dx/dy=1
f(x)=2x+cos(x) and g=f^-1. Find g'(1)

InteGrand · Mar 5, 2016

leehuan said:
More stupid questions... I should just chill with the maths for tonight but.

1. How would I prove that a function is injective and surjective? Trying to prove that f(x)=x^3+3x+1 has an inverse over \mathbb R. (Without graphical approach)

All I thought about is f'(x)=3x^2+3 is positive so f(x) monotonic increasing. Is that enough?
____________

And

2. Forgot how to use dy/dx.dx/dy=1
f(x)=2x+cos(x) and g=f^-1. Find g'(1)

$\noindent For 1, let's assume we have defined the domain and codomain as $\mathbb{R}$. It is injective as it is monotonic, as you proved. It is also surjective, because given any $y$ in the codomain ($\mathbb{R}$), there exists $x\in \mathbb{R}$ such that $f(x)=y$ (since this is a cubic equation (with real coefficients), which we know has a real root always; this fact follows from the Intermediate Value Theorem. In fact, it is similarly true that any polynomial equation of odd degree with only real coefficients will have a real root).$

$\noindent 2. Note that $f^\prime (x) = 2 -\sin x$. Also note that $f(0) = 1$, so by the inverse function theorem (linked below for anyone curious), we have$

$\begin{align*}g^\prime (1) &= \frac{1}{f^\prime (0)} \\ &= \frac{1}{2-\sin 0} \\ &= \frac{1}{2}.\end{align*}$

Inverse function theorem: https://en.wikipedia.org/wiki/Inverse_function_theorem#Statement_of_the_theorem .

Paradoxica · Mar 5, 2016

leehuan said:
More stupid questions... I should just chill with the maths for tonight but.

1. How would I prove that a function is injective and surjective? Trying to prove that f(x)=x^3+3x+1 has an inverse over \mathbb R. (Without graphical approach)

All I thought about is f'(x)=3x^2+3 is positive so f(x) monotonic increasing. Is that enough?
____________

And

2. Forgot how to use dy/dx.dx/dy=1
f(x)=2x+cos(x) and g=f^-1. Find g'(1)

Check out what I did in the Elementary Thread.

A function is injective if and only if

$$\noindent$f(x) = f(y) \Leftrightarrow x=y\\$Examples:$\\f(x) = x^2 \\ f(x)=f(y) \Leftrightarrow x^2 = y^2 \Leftrightarrow x = \pm y \\ $So $f(x) = x^2$ is NOT injective. But we already knew that.$\\$Often times it is not straightforward to see how you could get $x=y$ from $f(x) = f(y)$, but if you do manage to do it, then you know for certain the function is injective.$

leehuan · Mar 6, 2016

Oh crap. THAT was how to use the inverse function theorem. Ok I will sleep to recover.

Thanks again to responders.

Also, @Para the statement that a function is injective iff f(x)=f(y)<=>x=y - I have seen it in my lecture. But I kept scratching my head cause I'm not sure why it works.

InteGrand · Mar 6, 2016

leehuan said:
Oh crap. THAT was how to use the inverse function theorem. Ok I will sleep to recover.

Thanks again to responders.

Also, @Para the statement that a function is injective iff f(x)=f(y)<=>x=y - I have seen it in my lecture. But I kept scratching my head cause I'm not sure why it works.

$\noindent It means that the definition of a function being injective is that $f(x) = f(y)$ implies $x=y$. (Don't worry about the $\Leftarrow$ part, that's obvious by definition of any function, not just injective ones. That just says if we sub. in the same value to the function, the output is the same, which is obvious.)$

$\noindent So basically it's saying that, for an injective function (one-to-one), if the function attains the same value at ``two'' places (i.e. $f(x) = f(y)$), it must be the case that these two places are actually the same, i.e. it implies $x=y$.$

$\noindent As an example, say we're considering the function $f:\left[0,\infty) \to \mathbb{R}$ given by $f(x) = \sqrt{x}$. (We know this function is one-to-one.). This function satisfies the property that if $\sqrt{x} = \sqrt{y}$, then $x=y$ (in other words, if two non-negative numbers have the same square root, they are in fact the same number. This is in contrast to say $g:\mathbb{R} \to \mathbb{R}$ given by $g(x) =\sin x$. Just because two numbers have the same sine, it doesn't mean they're equal. In other words, $g(x) = g(y)$ does NOT imply that $x=y$, so this function is NOT one-to-one (as we knew).).$

$\noindent (So the reason that the statement ``a one-to-one function $f: X\to Y$ satisfies $f(x_1) = f(x_2) \Rightarrow x_1=x_2$ for all $x_1,x_2 \in X$'' works is that it is simply the definition of ``one-to-one''.)$

Paradoxica · Mar 6, 2016

leehuan said:
Oh crap. THAT was how to use the inverse function theorem. Ok I will sleep to recover.

Thanks again to responders.

Also, @Para the statement that a function is injective iff f(x)=f(y)<=>x=y - I have seen it in my lecture. But I kept scratching my head cause I'm not sure why it works.

Because in an injective function, every element in the domain and co-domain can only occur once. Thus, every pairing is unique. So f(x) = f(y) is logically equivalent to x=y, otherwise, that would contradict the definition of an injective function.

leehuan · Mar 6, 2016

Oh my goodness... of course... I am so foolish

Paradoxica · Mar 6, 2016

leehuan said:
Oh my goodness... of course... I am so foolish

We've all been there. Probably.

leehuan · Mar 6, 2016

Followup to Q1.

$\\ $Find all possible intervals $A\subseteq\mathbb R$, as large as possible, such that the function defined on $A$ by $f(x)={x}^{3}-3x+1$ has an inverse. What is the domain of each of these inverse functions?$

This just sounds like HSC MX1. I can assume f is surjective like above and then use my knowledge of turning points to determine intervals that allow it to be injective as well right?

seanieg89 · Mar 6, 2016

Yep exactly. Because we want our domain to be an interval, the turning points of the cubic are "bad" in the sense that f is not injective in any inteval containing either of these turning points in their interior.

Elsewhere we don't have a problem, because f is piecewise monotonic, so the intervals are precisely the three intervals that these two turning points cut the real line into.

(And then just apply f to each of these intervals to get the domains for the respective inverse functions.)

leehuan · Mar 6, 2016

Awesome. I don't suck too much then.

Aside: My syllabus is somewhere in the middle of here if any helper needs to know my progress
https://www.maths.unsw.edu.au/sites/default/files/math11512016.pdf

leehuan · Mar 6, 2016

$\\ $RTP: $\arcsin{x}+\arccos{x}$ is a constant. \\ All I want to know if there's a more elegant approach to taking derivatives and then reversing the differentiation process using integration?$$

Followup: Obviously arctan(x)+arccot(x) becomes a piecewise function. How do I justify that this isn't the case for arcsin(x)+arccos(x)? Does it basically just have to do with the fact arccot(x) is undefined for x=0 but continuous everywhere else?

Paradoxica · Mar 6, 2016

leehuan said:
$\\ $RTP: $\arcsin{x}+\arccos{x}$ is a constant. \\ All I want to know if there's a more elegant approach to taking derivatives and then reversing the differentiation process using integration?$$

Followup: Obviously arctan(x)+arccot(x) becomes a piecewise function. How do I justify that this isn't the case for arcsin(x)+arccos(x)? Does it basically just have to do with the fact arccot(x) is undefined for x=0 but continuous everywhere else?

Um... Complementary angles prove it by definition...

As for arccot, it satisfies the complementary angles, but two different cases for positive and negative branches.

leehuan · Mar 6, 2016

Paradoxica said:
Um... Complementary angles prove it by definition...

As for arccot, it satisfies the complementary angles, but two different cases for positive and negative branches.

Wow. I thought I was missing something but that was bad from me.

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