mathmath333 one year ago find the domain and the range of the function

1. mathmath333

\large \color{black}{\begin{align}f(x)=\dfrac{x^2+2x+1}{x^2-8x+12}\hspace{.33em}\\~\\ \end{align}}

2. acxbox22

factor the top and the bottom please

3. mathmath333

\large \color{black}{\begin{align} f(x)=\dfrac{(x+1)^2}{(x-6)(x-2)}\hspace{.33em}\\~\\ \end{align}}

4. acxbox22

no forget that...that doesn't really help

5. anonymous

for domain x^2 -8x +12 not equal to zero on factorization (x-6)(x-2) so domain R-{6,2} for range write f(x) as y, cross multiply it and form a quadratic of x D>=0 you will get the range of y which is the range

6. mathmath333

what is D>=0

7. anonymous

discriminant of the quadratic function greater than or equal to zero coz we are using x as real so D

8. mathmath333

so in every case do u have to take D>=0 , or it can also be taken as D<=0 or D=0

9. anonymous

no as we are taking x as a real value we have to take D>=0 if we take D<=0 that would mean that x is taking as a complex number that would make no sense of range

10. xapproachesinfinity

i don't think that range can be achieved with regular methods need calculus

11. mathmath333

Range all reals except -21/4<y<0 How to write this in notation form

12. xapproachesinfinity

(-21/4, 0) set builder notation

13. xapproachesinfinity

how did you find the range, by the described method above ?

14. mathmath333

r domain x^2 -8x +12 not equal to zero on factorization (x-6)(x-2) so domain R-{6,2} for range write f(x) as y, cross multiply it and form a quadratic of x D>=0 you will get the range of y which is the range

15. xapproachesinfinity

or you meant {y in R| -21/4<y<0}

16. mathmath333

u have to take the discriminant from there

17. xapproachesinfinity

let me see the graph

18. xapproachesinfinity

http://prntscr.com/7biuyn does not seem to be thaat your range is correct

19. xapproachesinfinity

(-oo, -5 1/4]u[0, oo)

20. xapproachesinfinity

according to the graph

21. mathmath333
22. xapproachesinfinity

yeah that's the same

23. xapproachesinfinity

24. mathmath333

25. xapproachesinfinity

oh isee my bad i meant -5 and 1/4

26. xapproachesinfinity

and -1/4 not 1/4 lol

27. xapproachesinfinity

so -21/4 in other words

28. xapproachesinfinity

i think that be achieved true calculus not regular methods

29. xapproachesinfinity

30. mathmath333

its same as i mentioned \large \color{black}{\begin{align} \mathbb{R}-\{(-\dfrac{21}{4},0)\}\hspace{.33em}\\~\\ \end{align}} and i never learned calculus

31. xapproachesinfinity

not really the way you just wrote is mean -21/4 and 0 are discarded but those are part of the range in our case

32. xapproachesinfinity

the correct way in set builder notation is the way wolfram wrote it

33. mathmath333

u mean it should be this \large \color{black}{\begin{align} \mathbb{R}-\{[-\dfrac{21}{4},0]\}\hspace{.33em}\\~\\ \end{align}}

34. xapproachesinfinity

here it is not a point that is discarded but a portion of real number from -21/4 to 0 but not including the those points i see you got the right thing, but missed a little bit when i read your reply first time i thought that interval was your range

35. xapproachesinfinity

eh no just miss understood your notation it was correct the first time

36. xapproachesinfinity

that was correct :)

37. xapproachesinfinity

how did you get it though?

38. mathmath333

this is corrrect below \large \color{black}{\begin{align} \mathbb{R}-\{(-\dfrac{21}{4},0)\}\hspace{.33em}\\~\\ \end{align}}

39. xapproachesinfinity

yes!

40. mathmath333

finnallly!!

41. xapproachesinfinity

you found discriminant?

42. mathmath333

yep

43. xapproachesinfinity

hmm i see, i must be wrong thinking it is not doable lol

44. xapproachesinfinity

hmm i see, now i remember what you did

45. mathmath333

by replacing $$f(x)$$ by $$y$$ u will get , \large \color{black}{\begin{align} (y-1)x^2-x(8y+2)=1\hspace{.33em}\\~\\ \end{align}}

46. xapproachesinfinity

yeah i just saw it!

47. mathmath333

their is a phrase "came late but came fit"

48. xapproachesinfinity

true :) thanks i learned something that i forgot from this!

49. xapproachesinfinity

i would usually throw my into calculus concepts to find a way to do this

50. xapproachesinfinity

the co-domain stuff are important in real analysis later if you are doing maths

51. mathmath333

not this far for me

52. xapproachesinfinity

i see, well you seem to like math a lot, so enjoy it

53. ybarrap

So to summarize what you did: We replace f(x) with y and solve the following for x $$x^2 (y-1)-x (8 y+2) = 1-12 y$$ If y=1 then x=11/10, so y=1 is in the range of f(x). For $$y\ne1$$ $$x =\cfrac{ 4 y-\sqrt{y (4 y+21)}+1}{y-1}$$ For real x, we need $$\sqrt{y (4 y+21)}\ge0\\ \implies y\ge0\text{ & }4 y+21\ge 0\text{ or }y\ge\cfrac{-21}{4}\\$$ The intersection of $$y\ge0$$ and $$y\ge\cfrac{-21}{4}$$ is $$y\ge0$$ But also $$\sqrt{y (4 y+21)}\ge0\\ \implies y\le0\text{ & }4 y+21\le 0\text{ or }y\le\cfrac{-21}{4}$$ So the range of $$f(x)=y$$ is $$\text{(}y\text{|}y\le\cfrac{-21}{4}\cup y\ge0\text{)}$$