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mathmath333

  • one year ago

find the domain and the range of the function

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}f(x)=\dfrac{x^2+2x+1}{x^2-8x+12}\hspace{.33em}\\~\\ \end{align}}\)

  2. acxbox22
    • one year ago
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    factor the top and the bottom please

  3. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} f(x)=\dfrac{(x+1)^2}{(x-6)(x-2)}\hspace{.33em}\\~\\ \end{align}}\)

  4. acxbox22
    • one year ago
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    no forget that...that doesn't really help

  5. anonymous
    • one year ago
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    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
    • one year ago
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    what is D>=0

  7. anonymous
    • one year ago
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    discriminant of the quadratic function greater than or equal to zero coz we are using x as real so D

  8. mathmath333
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    i don't think that range can be achieved with regular methods need calculus

  11. mathmath333
    • one year ago
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    Range all reals except -21/4<y<0 How to write this in notation form

  12. xapproachesinfinity
    • one year ago
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    (-21/4, 0) set builder notation

  13. xapproachesinfinity
    • one year ago
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    how did you find the range, by the described method above ?

  14. mathmath333
    • one year ago
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    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
    • one year ago
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    or you meant {y in R| -21/4<y<0}

  16. mathmath333
    • one year ago
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    u have to take the discriminant from there

  17. xapproachesinfinity
    • one year ago
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    let me see the graph

  18. xapproachesinfinity
    • one year ago
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    http://prntscr.com/7biuyn does not seem to be thaat your range is correct

  19. xapproachesinfinity
    • one year ago
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    (-oo, -5 1/4]u[0, oo)

  20. xapproachesinfinity
    • one year ago
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    according to the graph

  21. xapproachesinfinity
    • one year ago
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    yeah that's the same

  22. xapproachesinfinity
    • one year ago
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    your range missed a lot

  23. mathmath333
    • one year ago
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    i read it as -51/4

  24. xapproachesinfinity
    • one year ago
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    oh isee my bad i meant -5 and 1/4

  25. xapproachesinfinity
    • one year ago
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    and -1/4 not 1/4 lol

  26. xapproachesinfinity
    • one year ago
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    so -21/4 in other words

  27. xapproachesinfinity
    • one year ago
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    i think that be achieved true calculus not regular methods

  28. xapproachesinfinity
    • one year ago
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    through* instead true

  29. mathmath333
    • one year ago
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    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

  30. xapproachesinfinity
    • one year ago
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    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

  31. xapproachesinfinity
    • one year ago
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    the correct way in set builder notation is the way wolfram wrote it

  32. mathmath333
    • one year ago
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    u mean it should be this \(\large \color{black}{\begin{align} \mathbb{R}-\{[-\dfrac{21}{4},0]\}\hspace{.33em}\\~\\ \end{align}}\)

  33. xapproachesinfinity
    • one year ago
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    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

  34. xapproachesinfinity
    • one year ago
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    eh no just miss understood your notation it was correct the first time

  35. xapproachesinfinity
    • one year ago
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    that was correct :)

  36. xapproachesinfinity
    • one year ago
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    how did you get it though?

  37. mathmath333
    • one year ago
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    this is corrrect below \(\large \color{black}{\begin{align} \mathbb{R}-\{(-\dfrac{21}{4},0)\}\hspace{.33em}\\~\\ \end{align}}\)

  38. xapproachesinfinity
    • one year ago
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    yes!

  39. mathmath333
    • one year ago
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    finnallly!!

  40. xapproachesinfinity
    • one year ago
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    you found discriminant?

  41. mathmath333
    • one year ago
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    yep

  42. xapproachesinfinity
    • one year ago
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    hmm i see, i must be wrong thinking it is not doable lol

  43. xapproachesinfinity
    • one year ago
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    hmm i see, now i remember what you did

  44. mathmath333
    • one year ago
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    by replacing \(f(x)\) by \(y\) u will get , \(\large \color{black}{\begin{align} (y-1)x^2-x(8y+2)=1\hspace{.33em}\\~\\ \end{align}}\)

  45. xapproachesinfinity
    • one year ago
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    yeah i just saw it!

  46. mathmath333
    • one year ago
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    their is a phrase "came late but came fit"

  47. xapproachesinfinity
    • one year ago
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    true :) thanks i learned something that i forgot from this!

  48. xapproachesinfinity
    • one year ago
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    i would usually throw my into calculus concepts to find a way to do this

  49. xapproachesinfinity
    • one year ago
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    the co-domain stuff are important in real analysis later if you are doing maths

  50. mathmath333
    • one year ago
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    not this far for me

  51. xapproachesinfinity
    • one year ago
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    i see, well you seem to like math a lot, so enjoy it

  52. ybarrap
    • one year ago
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    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 \(\text{(}y\text{|}y\le\cfrac{-21}{4}\cup y\ge0\text{)}\)

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