find the domain and the range of the function

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- mathmath333

find the domain and the range of the function

- jamiebookeater

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- mathmath333

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

- acxbox22

factor the top and the bottom please

- mathmath333

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- acxbox22

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

- 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

what is D>=0

- anonymous

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

- 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

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

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

- mathmath333

Range all reals except -21/4

- xapproachesinfinity

(-21/4, 0)
set builder notation

- xapproachesinfinity

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

- 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

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

- mathmath333

u have to take the discriminant from there

- xapproachesinfinity

let me see the graph

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

according to the graph

- 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

yeah that's the same

- xapproachesinfinity

your range missed a lot

- mathmath333

i read it as -51/4

- xapproachesinfinity

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

- xapproachesinfinity

and -1/4 not 1/4 lol

- xapproachesinfinity

so -21/4 in other words

- xapproachesinfinity

i think that be achieved true calculus not regular methods

- xapproachesinfinity

through* instead true

- 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

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

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

- mathmath333

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

- 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

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

- xapproachesinfinity

that was correct :)

- xapproachesinfinity

how did you get it though?

- mathmath333

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

- xapproachesinfinity

yes!

- mathmath333

finnallly!!

- xapproachesinfinity

you found discriminant?

- mathmath333

yep

- xapproachesinfinity

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

- xapproachesinfinity

hmm i see, now i remember what you did

- 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

yeah i just saw it!

- mathmath333

their is a phrase "came late but came fit"

- xapproachesinfinity

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

- xapproachesinfinity

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

- xapproachesinfinity

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

- mathmath333

not this far for me

- xapproachesinfinity

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

- 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 \(\text{(}y\text{|}y\le\cfrac{-21}{4}\cup y\ge0\text{)}\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.