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anonymous
 one year ago
Determine if the following functions are bijective , and if so case determine the inverse:
anonymous
 one year ago
Determine if the following functions are bijective , and if so case determine the inverse:

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.01) \[f(x)=x \sqrt{2+x^2}\] 2) \[f(x)=xx+1\] 3) \[f(x)=\]dw:1444147581351:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix you know how to do this?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0\[\large\rm f(x)=x\sqrt{2+x^2}\] Notice that the x is being squared under the root. So the stuff under the root is always positive. We have `no restrictions` on our inputs for x. So our domain is \(\large\rm \mathbb R\) ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0And the range is also all real numbers. So the function is onto. The entire range is covered. I can't remember how to show onto >.< Maybe we can just say that... Domain = all real numbers Range = all real numbers so f(x) is onto.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Onetoone is a little easier to check. You need to be able to pass the `horizontal line test` in order to be onetoone.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0We can do that by looking at \(\large\rm f(x)\).  Even functions have this property:\[\large\rm f(x)=f(x)\]and are not onetoone.  Odd functions however have this property:\[\large\rm f(x)=f(x)\]and are onetoone.  So with our function, what happens when you replace x with x? \[\large\rm f(\color{orangered}{x})=(\color{orangered}{x})\sqrt{2+(\color{orangered}{x})^2}\]\[\large\rm f(\color{royalblue}{x})=(\color{royalblue}{x})\sqrt{2+(\color{royalblue}{x})^2}\]What is that going to simplify to? :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the one with x is going to be negative? and if it would become positive we would have an one to one? or is the horizontal test that it will be the exact same but on the negative yaxis?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0The one with the negative x is going to be negative. It's the `exact same equation` but with a negative sign in front. (Make sure you understand that the negative gets squared away under the root.)\[\large\rm f(\color{royalblue}{x})=x\sqrt{2+x^2}\]This is the function we started with but with a negative in front,\[\large\rm f(x)=\color{royalblue}{x\sqrt{2+x^2}}\]\[\large\rm f(x)=\color{royalblue}{f(x)}\]So we showed that our function has the property of an odd function, so yes, it is onetoone.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha yeah, it is!! so, when its onto and one to one its bijective, yeah?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Hmm I'm a little rusty on this stuff... I should be careful what I'm saying. Being an odd function doesn't automatically make it onetoone. sin(x) is a good example of that. We're ok on the first problem though :o Just thinking..

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Hmm sorry mr koala :c my brain being stupid today. lemme see if smart guy can help @ganeshie8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but isnt that one of the terms of being a bijective equation?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0isn't `that`? that what? :o

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Are we looking at a certain question above?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh sorry, isnt that one of the conditions i mean

freckles
 one year ago
Best ResponseYou've already chosen the best response.1If a function is onto and onetoone then it is bijective.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0looking at #1. I just been spittin a bunch of nonsense :D I don't recall how to show onetoone and onto lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[f(a)=f(b) \text{ gives only } a=b \text{ then it is one to one }\] We could try to show it is always increasing... \[f(x)=x \sqrt{2+x^2} \\ f'(x)=\sqrt{2+x^2}+x \frac{1}{2}(2x) \frac{1}{\sqrt{2+x^2}} \\ f'(x)=\sqrt{2+x^2}+\frac{x^2}{\sqrt{2+x^2}} \\ \\ \\ f'(x)=\frac{1}{\sqrt{2+x^2}}(2+x^2+x^2) \\ f'(x)=\frac{2+2x^2}{\sqrt{2+x^2}}\] \[f'>0 \text{ for all } x \text{ so } f \text{ is always increasing } \\ \text{ so } f \text{ is one to one }\] I tried to do it the other way but it looked really gross

freckles
 one year ago
Best ResponseYou've already chosen the best response.1now what are the codomains for which these function are defined?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[f:A>B \\ \text{ defined by } f(x)=2 \sqrt{2+x^2} \\ \text{ has codomain } B \\ \text{ To show it is surjective we must show }\\ \text{ for all } y \in B \text{ there exist } x \in A \\ \text{ such that } f(x)=y\]  other word for surjective is onto (might be other words but I used onto often)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1basically we just want to show every element of the codomain gets hit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and how do we show that?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I was asking about your codomain you need to answer that question first

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I don't know... It should be given.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1These questions area weird if they don't get you the domain and codomain

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Like it should say let f:A>B be defined by f(x)=something goes here

freckles
 one year ago
Best ResponseYou've already chosen the best response.1where A and B are sets numbers given

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so do you think we should assume both domain and codomain are given as real numbers @pate16 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, but it says under the assignment B=Vf for every case

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Well I don't know what that means either. But here is an example: \[\text{ Show } f: \mathbb{R}\rightarrow \mathbb{R} \text{ is defined by } f(x)=x^3 \text{ is bijective }\] \[\text{ to show it is injective we could do that whole derivative thing or try } \\ \text{ to do the whole } \\ f(a)=f(b) \text{ implies only } a=b\] \[f'(x)=3x^2>0 \text{ for all } x \text{ so } f \text{ is increasing on } \mathbb{R}\] \[f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by} f(x)=x^3 \text{ is } \\ \text{ injective } \\ \text{ OR } \\ f(a)=f(b) \\ a^3=b^3 \\ a^3b^3=0 \\ (ab)(a^2+ab+b^2)=0 \\ a=b \text{ or } a=\frac{b \pm \sqrt{b^2 4b^2}}{2}=\frac{b \pm \sqrt{3b^2}}{2} \\ \text{ second option doesn't matter anyways since we are only looking } \\ \text{ at the set of real numbers } \\ \text{ so we do have } f(a)=f(b) \implies a=b \] \[\text{ surjective } \\ \text{ we can just pick an element of the domain to show we can get } \\ \text{ every element of the codomain }\] \[\sqrt[3]{b} \in \mathbb{R} \\ f( \sqrt[3]{b})=(\sqrt[3]{b})^3=b \in \mathbb{R}\] \[\text{ So } f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } f(x)=x^3 \text{ onto } \mathbb{R} \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[g: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } g(x)=x^2 \text{ is not surjective on } \mathbb{R} \text{ because } 1 \in \mathbb{R} \text{ but } \\ x^2 \neq 1 \text{ for any real } x \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I hope this shows why we need to know the domain and codomain

freckles
 one year ago
Best ResponseYou've already chosen the best response.1like if I said \[g: \mathbb{R} \rightarrow [0,\infty) \text{ defined by } g(x)=x^2 \text{ this would be surjective over the } [0,\infty)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1let's assume \[\text{ we have } \\ f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } f(x)=x \sqrt{2+x^2} \\ \text{ we need to show for every } b \in \mathbb{R} \text{ there exist } a \in \mathbb{R} \\ \text{ such that } f(a)=b\] so first we must go backwards before we jump into the actual proof \[f(a)=a \sqrt{2+a^2}=b \\ a \sqrt{2+a^2}=b \\ a^2(2+a^2)=b^2 \\ a^4+2a^2b^2=0 \\ a^2=\frac{ 2 \pm \sqrt{4+4b^2}}{2} \\ a^2=1 \pm \sqrt{1+b^2} \\ \text{ choosing } a^2=1+\sqrt{1+b^2} \\ \text{ since the other one says } a^2 \text{ is a negative number } \\ a = \pm \sqrt{1 + \sqrt{1+b^2}}\] So...for the proof... \[\sqrt{1+\sqrt{1+b^2}},\in \mathbb{R} \\ f(\sqrt{1 + \sqrt{1+b^2}})=\sqrt{1+\sqrt{1+b^2}} \sqrt{2+(\sqrt{1+\sqrt{1+b^2}})^2} \\ = \sqrt{1+\sqrt{1+b^2}} \sqrt{21+\sqrt{1+b^2}} \\ =\sqrt{\sqrt{1+b^2}1} \sqrt{\sqrt{1+b^2}+1} \\ =\sqrt{(1+b^2)1)} \\ =\sqrt{b^2}=b \text{ if } b>0 \text{ so we have shown we can get every } b \text{ in the codomain } \\ \text{ therefore } f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } f(x)=x \sqrt{2+x^2} \text{ is surjective over the reals }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you can show in a similar way the negative reals also get hit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okey, so this is the tactics for all of these ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I made assumptions what the domain and codomain are... but yeah I'm just applying the definitions of injective and surjective however I did give another way to show a function is injective which I think makes some of these easier

freckles
 one year ago
Best ResponseYou've already chosen the best response.1You can just look at a graph though if you are just to suppose to determine and you don't really need a proof but you still need to know what domain and codomain to look at

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is hard stuff..

freckles
 one year ago
Best ResponseYou've already chosen the best response.1another example: dw:1444153993613:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444154104419:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } f(x)=x \sqrt{2+x^2} \text{ looks something like this }\] dw:1444154363592:dw horrible drawing but we get the idea that every y is going to get hit and every y gets hit only once

freckles
 one year ago
Best ResponseYou've already chosen the best response.1where y is a number of the codomain

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444154427501:dw
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