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

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2\(\large \color{black}{\begin{align}f(x)=\dfrac{x^2+2x+1}{x^28x+12}\hspace{.33em}\\~\\ \end{align}}\)

acxbox22
 one year ago
Best ResponseYou've already chosen the best response.0factor the top and the bottom please

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2\(\large \color{black}{\begin{align} f(x)=\dfrac{(x+1)^2}{(x6)(x2)}\hspace{.33em}\\~\\ \end{align}}\)

acxbox22
 one year ago
Best ResponseYou've already chosen the best response.0no forget that...that doesn't really help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for domain x^2 8x +12 not equal to zero on factorization (x6)(x2) 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0discriminant of the quadratic function greater than or equal to zero coz we are using x as real so D

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2so in every case do u have to take D>=0 , or it can also be taken as D<=0 or D=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no 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

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i don't think that range can be achieved with regular methods need calculus

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2Range all reals except 21/4<y<0 How to write this in notation form

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0(21/4, 0) set builder notation

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0how did you find the range, by the described method above ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2r domain x^2 8x +12 not equal to zero on factorization (x6)(x2) 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

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0or you meant {y in R 21/4<y<0}

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2u have to take the discriminant from there

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0let me see the graph

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0http://prntscr.com/7biuyn does not seem to be thaat your range is correct

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0(oo, 5 1/4]u[0, oo)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0according to the graph

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0yeah that's the same

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0your range missed a lot

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2i read it as 51/4

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0oh isee my bad i meant 5 and 1/4

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0and 1/4 not 1/4 lol

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0so 21/4 in other words

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i think that be achieved true calculus not regular methods

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0through* instead true

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2its same as i mentioned \(\large \color{black}{\begin{align} \mathbb{R}\{(\dfrac{21}{4},0)\}\hspace{.33em}\\~\\ \end{align}}\) and i never learned calculus

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0not really the way you just wrote is mean 21/4 and 0 are discarded but those are part of the range in our case

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0the correct way in set builder notation is the way wolfram wrote it

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2u mean it should be this \(\large \color{black}{\begin{align} \mathbb{R}\{[\dfrac{21}{4},0]\}\hspace{.33em}\\~\\ \end{align}}\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0here 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

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0eh no just miss understood your notation it was correct the first time

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0that was correct :)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0how did you get it though?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2this is corrrect below \(\large \color{black}{\begin{align} \mathbb{R}\{(\dfrac{21}{4},0)\}\hspace{.33em}\\~\\ \end{align}}\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0you found discriminant?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0hmm i see, i must be wrong thinking it is not doable lol

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0hmm i see, now i remember what you did

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2by replacing \(f(x)\) by \(y\) u will get , \(\large \color{black}{\begin{align} (y1)x^2x(8y+2)=1\hspace{.33em}\\~\\ \end{align}}\)

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0yeah i just saw it!

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2their is a phrase "came late but came fit"

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0true :) thanks i learned something that i forgot from this!

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i would usually throw my into calculus concepts to find a way to do this

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0the codomain stuff are important in real analysis later if you are doing maths

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2not this far for me

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.0i see, well you seem to like math a lot, so enjoy it

ybarrap
 one year ago
Best ResponseYou've already chosen the best response.0So to summarize what you did: We replace f(x) with y and solve the following for x $$ x^2 (y1)x (8 y+2) = 112 y $$ If y=1 then x=11/10, so y=1 is in the range of f(x). For \(y\ne1\) $$ x =\cfrac{ 4 y\sqrt{y (4 y+21)}+1}{y1} $$ 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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