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

1. anonymous

1) $f(x)=x \sqrt{2+x^2}$ 2) $f(x)=x|x|+1$ 3) $f(x)=$||dw:1444147581351:dw|

2. anonymous

@zepdrix you know how to do this?

3. zepdrix

So ummm

4. 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?

5. 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.

6. 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.

7. 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? :)

8. 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?

9. 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.

10. zepdrix

Confusing stuff? :d

11. zepdrix

hmm

12. anonymous

haha yeah, it is!! so, when its onto and one to one its bijective, yeah?

13. zepdrix

yes c:

14. 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..

15. zepdrix

Hmm sorry mr koala :c my brain being stupid today. lemme see if smart guy can help @ganeshie8

16. zepdrix

@freckles

17. anonymous

but isnt that one of the terms of being a bijective equation?

18. zepdrix

isn't that? that what? :o

19. freckles

Are we looking at a certain question above?

20. anonymous

oh sorry, isnt that one of the conditions i mean

21. freckles

If a function is onto and one-to-one then it is bijective.

22. 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

23. 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

24. freckles

now what are the codomains for which these function are defined?

25. 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)

26. freckles

basically we just want to show every element of the codomain gets hit

27. anonymous

and how do we show that?

28. freckles

29. anonymous

it is B? or what?

30. freckles

I don't know... It should be given.

31. freckles

These questions area weird if they don't get you the domain and codomain

32. freckles

Like it should say let f:A->B be defined by f(x)=something goes here

33. anonymous

B=Vf

34. freckles

where A and B are sets numbers given

35. freckles

so do you think we should assume both domain and codomain are given as real numbers @pate16 ?

36. anonymous

yeah, but it says under the assignment B=Vf for every case

37. freckles

what does that mean

38. anonymous

value amount

39. freckles

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

40. freckles

$g: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } g(x)=x^2 \text{ is not surjective on } \mathbb{R} \text{ because } -1 \in \mathbb{R} \text{ but } \\ x^2 \neq -1 \text{ for any real } x$

41. freckles

I hope this shows why we need to know the domain and codomain

42. freckles

like if I said $g: \mathbb{R} \rightarrow [0,\infty) \text{ defined by } g(x)=x^2 \text{ this would be surjective over the } [0,\infty)$

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

44. freckles

you can show in a similar way the negative reals also get hit

45. anonymous

Okey, so this is the tactics for all of these ?

46. 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

47. 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

48. anonymous

this is hard stuff..

49. freckles

another example: |dw:1444153993613:dw|

50. freckles

|dw:1444154104419:dw|

51. 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

52. freckles

where y is a number of the codomain

53. freckles

|dw:1444154427501:dw|