Question

find the domain and the range of the function

Answer 1

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

Answer 2

factor the top and the bottom please

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

what is D>=0

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Range all reals except -21/4<y<0

How to write this in notation form

Answer 12

(-21/4, 0)
set builder notation

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

u have to take the discriminant from there

Answer 17

let me see the graph

Answer 18

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

Answer 19

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

Answer 20

according to the graph

Answer 21

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

Answer 22

yeah that's the same

Answer 23

your range missed a lot

Answer 24

i read it as -51/4

Answer 25

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

Answer 26

and -1/4 not 1/4 lol

Answer 27

so -21/4 in other words

Answer 28

i think that be achieved true calculus not regular methods

Answer 29

through* instead true

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

u mean it should be this

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

Answer 34

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

Answer 35

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

Answer 36

that was correct :)

Answer 37

how did you get it though?

Answer 38

this is corrrect below

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

Answer 39

yes!

Answer 40

finnallly!!

Answer 41

you found discriminant?

Answer 42

yep

Answer 43

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

Answer 44

hmm i see, now i remember what you did

Answer 45

by replacing \(f(x)\) by \(y\) u will get ,


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

Answer 46

yeah i just saw it!

Answer 47

their is a phrase "came late but came fit"

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

not this far for me

Answer 52

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

Answer 53

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