mathmath333
  • mathmath333
find the domain and the range of the function
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align}f(x)=\dfrac{x^2+2x+1}{x^2-8x+12}\hspace{.33em}\\~\\ \end{align}}\)
acxbox22
  • acxbox22
factor the top and the bottom please
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} f(x)=\dfrac{(x+1)^2}{(x-6)(x-2)}\hspace{.33em}\\~\\ \end{align}}\)

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acxbox22
  • acxbox22
no forget that...that doesn't really help
anonymous
  • 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
mathmath333
  • mathmath333
what is D>=0
anonymous
  • anonymous
discriminant of the quadratic function greater than or equal to zero coz we are using x as real so D
mathmath333
  • mathmath333
so in every case do u have to take D>=0 , or it can also be taken as D<=0 or D=0
anonymous
  • 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
xapproachesinfinity
  • xapproachesinfinity
i don't think that range can be achieved with regular methods need calculus
mathmath333
  • mathmath333
Range all reals except -21/4
xapproachesinfinity
  • xapproachesinfinity
(-21/4, 0) set builder notation
xapproachesinfinity
  • xapproachesinfinity
how did you find the range, by the described method above ?
mathmath333
  • 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
xapproachesinfinity
  • xapproachesinfinity
or you meant {y in R| -21/4
mathmath333
  • mathmath333
u have to take the discriminant from there
xapproachesinfinity
  • xapproachesinfinity
let me see the graph
xapproachesinfinity
  • xapproachesinfinity
http://prntscr.com/7biuyn does not seem to be thaat your range is correct
xapproachesinfinity
  • xapproachesinfinity
(-oo, -5 1/4]u[0, oo)
xapproachesinfinity
  • xapproachesinfinity
according to the graph
mathmath333
  • mathmath333
http://www.wolframalpha.com/input/?i=range+f%28x%29%3D%28%28x%2B1%29%5E2%29%2F%28%28x-2%29%28x-6%29%29
xapproachesinfinity
  • xapproachesinfinity
yeah that's the same
xapproachesinfinity
  • xapproachesinfinity
your range missed a lot
mathmath333
  • mathmath333
i read it as -51/4
xapproachesinfinity
  • xapproachesinfinity
oh isee my bad i meant -5 and 1/4
xapproachesinfinity
  • xapproachesinfinity
and -1/4 not 1/4 lol
xapproachesinfinity
  • xapproachesinfinity
so -21/4 in other words
xapproachesinfinity
  • xapproachesinfinity
i think that be achieved true calculus not regular methods
xapproachesinfinity
  • xapproachesinfinity
through* instead true
mathmath333
  • 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
xapproachesinfinity
  • 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
xapproachesinfinity
  • xapproachesinfinity
the correct way in set builder notation is the way wolfram wrote it
mathmath333
  • mathmath333
u mean it should be this \(\large \color{black}{\begin{align} \mathbb{R}-\{[-\dfrac{21}{4},0]\}\hspace{.33em}\\~\\ \end{align}}\)
xapproachesinfinity
  • 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
xapproachesinfinity
  • xapproachesinfinity
eh no just miss understood your notation it was correct the first time
xapproachesinfinity
  • xapproachesinfinity
that was correct :)
xapproachesinfinity
  • xapproachesinfinity
how did you get it though?
mathmath333
  • mathmath333
this is corrrect below \(\large \color{black}{\begin{align} \mathbb{R}-\{(-\dfrac{21}{4},0)\}\hspace{.33em}\\~\\ \end{align}}\)
xapproachesinfinity
  • xapproachesinfinity
yes!
mathmath333
  • mathmath333
finnallly!!
xapproachesinfinity
  • xapproachesinfinity
you found discriminant?
mathmath333
  • mathmath333
yep
xapproachesinfinity
  • xapproachesinfinity
hmm i see, i must be wrong thinking it is not doable lol
xapproachesinfinity
  • xapproachesinfinity
hmm i see, now i remember what you did
mathmath333
  • mathmath333
by replacing \(f(x)\) by \(y\) u will get , \(\large \color{black}{\begin{align} (y-1)x^2-x(8y+2)=1\hspace{.33em}\\~\\ \end{align}}\)
xapproachesinfinity
  • xapproachesinfinity
yeah i just saw it!
mathmath333
  • mathmath333
their is a phrase "came late but came fit"
xapproachesinfinity
  • xapproachesinfinity
true :) thanks i learned something that i forgot from this!
xapproachesinfinity
  • xapproachesinfinity
i would usually throw my into calculus concepts to find a way to do this
xapproachesinfinity
  • xapproachesinfinity
the co-domain stuff are important in real analysis later if you are doing maths
mathmath333
  • mathmath333
not this far for me
xapproachesinfinity
  • xapproachesinfinity
i see, well you seem to like math a lot, so enjoy it
ybarrap
  • 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 \(\text{(}y\text{|}y\le\cfrac{-21}{4}\cup y\ge0\text{)}\)

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