anonymous
  • anonymous
Determine if the following functions are bijective , and if so case determine the inverse:
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
1) \[f(x)=x \sqrt{2+x^2}\] 2) \[f(x)=x|x|+1\] 3) \[f(x)=\]||dw:1444147581351:dw|
anonymous
  • anonymous
@zepdrix you know how to do this?
zepdrix
  • zepdrix
So ummm

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zepdrix
  • zepdrix
\[\large\rm f(x)=x\sqrt{2+x^2}\] 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
  • zepdrix
And 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
  • zepdrix
One-to-one is a little easier to check. You need to be able to pass the `horizontal line test` in order to be one-to-one.
zepdrix
  • zepdrix
We can do that by looking at \(\large\rm f(-x)\). ----------------------------------------- Even functions have this property:\[\large\rm f(-x)=f(x)\]and are not one-to-one. ----------------------------------------- Odd functions however have this property:\[\large\rm f(-x)=-f(x)\]and are one-to-one. ----------------------------------------- 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
  • anonymous
the 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 y-axis?
zepdrix
  • zepdrix
The 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 one-to-one.
zepdrix
  • zepdrix
Confusing stuff? :d
zepdrix
  • zepdrix
hmm
anonymous
  • anonymous
haha yeah, it is!! so, when its onto and one to one its bijective, yeah?
zepdrix
  • zepdrix
yes c:
zepdrix
  • zepdrix
Hmm 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 one-to-one. sin(x) is a good example of that. We're ok on the first problem though :o Just thinking..
zepdrix
  • zepdrix
Hmm sorry mr koala :c my brain being stupid today. lemme see if smart guy can help @ganeshie8
zepdrix
  • zepdrix
@freckles
anonymous
  • anonymous
but isnt that one of the terms of being a bijective equation?
zepdrix
  • zepdrix
isn't `that`? that what? :o
freckles
  • freckles
Are we looking at a certain question above?
anonymous
  • anonymous
oh sorry, isnt that one of the conditions i mean
freckles
  • freckles
If a function is onto and one-to-one then it is bijective.
zepdrix
  • zepdrix
looking at #1. I just been spittin a bunch of nonsense :D I don't recall how to show one-to-one and onto lol
freckles
  • freckles
\[f(a)=f(b) \text{ gives only } a=b \text{ then it is one to one }\] 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
  • freckles
now what are the codomains for which these function are defined?
freckles
  • freckles
\[f:A->B \\ \text{ defined by } f(x)=2 \sqrt{2+x^2} \\ \text{ has codomain } B \\ \text{ To show it is surjective we must show }\\ \text{ for all } y \in B \text{ there exist } x \in A \\ \text{ such that } f(x)=y\] -- other word for surjective is onto (might be other words but I used onto often)
freckles
  • freckles
basically we just want to show every element of the codomain gets hit
anonymous
  • anonymous
and how do we show that?
freckles
  • freckles
I was asking about your codomain you need to answer that question first
anonymous
  • anonymous
it is B? or what?
freckles
  • freckles
I don't know... It should be given.
freckles
  • freckles
These questions area weird if they don't get you the domain and codomain
freckles
  • freckles
Like it should say let f:A->B be defined by f(x)=something goes here
anonymous
  • anonymous
B=Vf
freckles
  • freckles
where A and B are sets numbers given
freckles
  • freckles
so do you think we should assume both domain and codomain are given as real numbers @pate16 ?
anonymous
  • anonymous
yeah, but it says under the assignment B=Vf for every case
freckles
  • freckles
what does that mean
anonymous
  • anonymous
value amount
freckles
  • freckles
Well I don't know what that means either. But here is an example: \[\text{ Show } f: \mathbb{R}\rightarrow \mathbb{R} \text{ is defined by } f(x)=x^3 \text{ is bijective }\] \[\text{ to show it is injective we could do that whole derivative thing or try } \\ \text{ to do the whole } \\ f(a)=f(b) \text{ implies only } a=b\] \[f'(x)=3x^2>0 \text{ for all } x \text{ so } f \text{ is increasing on } \mathbb{R}\] \[f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by} f(x)=x^3 \text{ is } \\ \text{ injective } \\ \text{ OR } \\ f(a)=f(b) \\ a^3=b^3 \\ a^3-b^3=0 \\ (a-b)(a^2+ab+b^2)=0 \\ a=b \text{ or } a=\frac{-b \pm \sqrt{b^2 -4b^2}}{2}=\frac{-b \pm \sqrt{-3b^2}}{2} \\ \text{ second option doesn't matter anyways since we are only looking } \\ \text{ at the set of real numbers } \\ \text{ so we do have } f(a)=f(b) \implies a=b \] \[\text{ surjective } \\ \text{ we can just pick an element of the domain to show we can get } \\ \text{ every element of the codomain }\] \[\sqrt[3]{b} \in \mathbb{R} \\ f( \sqrt[3]{b})=(\sqrt[3]{b})^3=b \in \mathbb{R}\] \[\text{ So } f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } f(x)=x^3 \text{ onto } \mathbb{R} \]
freckles
  • freckles
\[g: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } g(x)=x^2 \text{ is not surjective on } \mathbb{R} \text{ because } -1 \in \mathbb{R} \text{ but } \\ x^2 \neq -1 \text{ for any real } x \]
freckles
  • freckles
I hope this shows why we need to know the domain and codomain
freckles
  • freckles
like 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
  • freckles
let'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 \[f(a)=a \sqrt{2+a^2}=b \\ a \sqrt{2+a^2}=b \\ a^2(2+a^2)=b^2 \\ a^4+2a^2-b^2=0 \\ a^2=\frac{- 2 \pm \sqrt{4+4b^2}}{2} \\ a^2=-1 \pm \sqrt{1+b^2} \\ \text{ choosing } a^2=-1+\sqrt{1+b^2} \\ \text{ since the other one says } a^2 \text{ is a negative number } \\ a = \pm \sqrt{-1 + \sqrt{1+b^2}}\] So...for the proof... \[\sqrt{-1+\sqrt{1+b^2}},\in \mathbb{R} \\ f(\sqrt{-1 + \sqrt{1+b^2}})=\sqrt{-1+\sqrt{1+b^2}} \sqrt{2+(\sqrt{-1+\sqrt{1+b^2}})^2} \\ = \sqrt{-1+\sqrt{1+b^2}} \sqrt{2-1+\sqrt{1+b^2}} \\ =\sqrt{\sqrt{1+b^2}-1} \sqrt{\sqrt{1+b^2}+1} \\ =\sqrt{(1+b^2)-1)} \\ =\sqrt{b^2}=b \text{ if } b>0 \text{ so we have shown we can get every } b \text{ in the codomain } \\ \text{ therefore } f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } f(x)=x \sqrt{2+x^2} \text{ is surjective over the reals }\]
freckles
  • freckles
you can show in a similar way the negative reals also get hit
anonymous
  • anonymous
Okey, so this is the tactics for all of these ?
freckles
  • freckles
I 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
  • freckles
You 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
  • anonymous
this is hard stuff..
freckles
  • freckles
another example: |dw:1444153993613:dw|
freckles
  • freckles
|dw:1444154104419:dw|
freckles
  • freckles
\[f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } f(x)=x \sqrt{2+x^2} \text{ looks something like this }\] |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
  • freckles
where y is a number of the codomain
freckles
  • freckles
|dw:1444154427501:dw|

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