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anonymous

  • one year ago

Determine if the following functions are bijective , and if so case determine the inverse:

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  1. anonymous
    • one year ago
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    1) \[f(x)=x \sqrt{2+x^2}\] 2) \[f(x)=x|x|+1\] 3) \[f(x)=\]||dw:1444147581351:dw|

  2. anonymous
    • one year ago
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    @zepdrix you know how to do this?

  3. zepdrix
    • one year ago
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    So ummm

  4. zepdrix
    • one year ago
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    \[\large\rm f(x)=x\sqrt{2+x^2}\] 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?

  5. zepdrix
    • one year ago
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    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.

  6. zepdrix
    • one year ago
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    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.

  7. zepdrix
    • one year ago
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    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? :)

  8. anonymous
    • one year ago
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    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?

  9. zepdrix
    • one year ago
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    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.

  10. zepdrix
    • one year ago
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    Confusing stuff? :d

  11. zepdrix
    • one year ago
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    hmm

  12. anonymous
    • one year ago
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    haha yeah, it is!! so, when its onto and one to one its bijective, yeah?

  13. zepdrix
    • one year ago
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    yes c:

  14. zepdrix
    • one year ago
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    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..

  15. zepdrix
    • one year ago
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    Hmm sorry mr koala :c my brain being stupid today. lemme see if smart guy can help @ganeshie8

  16. zepdrix
    • one year ago
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    @freckles

  17. anonymous
    • one year ago
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    but isnt that one of the terms of being a bijective equation?

  18. zepdrix
    • one year ago
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    isn't `that`? that what? :o

  19. freckles
    • one year ago
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    Are we looking at a certain question above?

  20. anonymous
    • one year ago
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    oh sorry, isnt that one of the conditions i mean

  21. freckles
    • one year ago
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    If a function is onto and one-to-one then it is bijective.

  22. zepdrix
    • one year ago
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    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

  23. freckles
    • one year ago
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    \[f(a)=f(b) \text{ gives only } a=b \text{ then it is one to one }\] 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

  24. freckles
    • one year ago
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    now what are the codomains for which these function are defined?

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

  26. freckles
    • one year ago
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    basically we just want to show every element of the codomain gets hit

  27. anonymous
    • one year ago
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    and how do we show that?

  28. freckles
    • one year ago
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    I was asking about your codomain you need to answer that question first

  29. anonymous
    • one year ago
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    it is B? or what?

  30. freckles
    • one year ago
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    I don't know... It should be given.

  31. freckles
    • one year ago
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    These questions area weird if they don't get you the domain and codomain

  32. freckles
    • one year ago
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    Like it should say let f:A->B be defined by f(x)=something goes here

  33. anonymous
    • one year ago
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    B=Vf

  34. freckles
    • one year ago
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    where A and B are sets numbers given

  35. freckles
    • one year ago
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    so do you think we should assume both domain and codomain are given as real numbers @pate16 ?

  36. anonymous
    • one year ago
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    yeah, but it says under the assignment B=Vf for every case

  37. freckles
    • one year ago
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    what does that mean

  38. anonymous
    • one year ago
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    value amount

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

  40. freckles
    • one year ago
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    \[g: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } g(x)=x^2 \text{ is not surjective on } \mathbb{R} \text{ because } -1 \in \mathbb{R} \text{ but } \\ x^2 \neq -1 \text{ for any real } x \]

  41. freckles
    • one year ago
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    I hope this shows why we need to know the domain and codomain

  42. freckles
    • one year ago
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    like if I said \[g: \mathbb{R} \rightarrow [0,\infty) \text{ defined by } g(x)=x^2 \text{ this would be surjective over the } [0,\infty)\]

  43. freckles
    • one year ago
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    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... \[\sqrt{-1+\sqrt{1+b^2}},\in \mathbb{R} \\ f(\sqrt{-1 + \sqrt{1+b^2}})=\sqrt{-1+\sqrt{1+b^2}} \sqrt{2+(\sqrt{-1+\sqrt{1+b^2}})^2} \\ = \sqrt{-1+\sqrt{1+b^2}} \sqrt{2-1+\sqrt{1+b^2}} \\ =\sqrt{\sqrt{1+b^2}-1} \sqrt{\sqrt{1+b^2}+1} \\ =\sqrt{(1+b^2)-1)} \\ =\sqrt{b^2}=b \text{ if } b>0 \text{ so we have shown we can get every } b \text{ in the codomain } \\ \text{ therefore } f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } f(x)=x \sqrt{2+x^2} \text{ is surjective over the reals }\]

  44. freckles
    • one year ago
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    you can show in a similar way the negative reals also get hit

  45. anonymous
    • one year ago
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    Okey, so this is the tactics for all of these ?

  46. freckles
    • one year ago
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    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

  47. freckles
    • one year ago
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    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

  48. anonymous
    • one year ago
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    this is hard stuff..

  49. freckles
    • one year ago
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    another example: |dw:1444153993613:dw|

  50. freckles
    • one year ago
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    |dw:1444154104419:dw|

  51. freckles
    • one year ago
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    \[f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } f(x)=x \sqrt{2+x^2} \text{ looks something like this }\] |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

  52. freckles
    • one year ago
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    where y is a number of the codomain

  53. freckles
    • one year ago
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    |dw:1444154427501:dw|

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