Question

Find the minimum and maximum value of

Answer 1

\(\large \color{black}{\begin{align} \dfrac{x^2-x+1}{x^2+x+1},\ x\in \mathbb{R}\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

It's hard to explain...but I will see if I can show the work.

Answer 3

are you talking local or absolute maxes and mins?

Answer 4

i would also prefer a way without calculus ( idk calculus)

Answer 5

Do you need to show your work for this problem?

Answer 6

not necessary i should understand the idea

Answer 7

Based on what I know algebraically , I got a minimum of (1, 1/3) and a maximum of (-1,3).

Answer 8

u calculated the co-ordinates where as it should be the numerical value

Answer 9

Ok. Let me try a different way.

Answer 10

Ok. I got the minimum is 1/3, and the maximum is -3.

Answer 11

oops +3. not neg

Answer 12

how did u calculate

Answer 13

\[\frac{x^2 - x + 1 }{ x^2 + x + 1 } = 3, when (x) = -1\]

I got this after simplifying to figure out what x = for the numerator

Answer 14

Same for the denominator, except it is when x=1

Answer 15

It's not the way I was taught, but my aunt is a math teacher, so she taught me her way of solving these pesky sorts

Answer 16

eh i like your questions lol

Answer 17

make me struggle :)

Answer 18

i have tons of question but no time

Answer 19

i cheated haha
min=1/3<y<3
let's how we get to this

Answer 20

one way seems to put random

\(x=\pm 0, \pm 1,\pm 2, \pm 3\) but doesnt work always

http://www.wolframalpha.com/input/?i=table%5Bf%28x%29%3D%5Cdfrac%7Bx%5E2-x%2B1%7D%7Bx%5E2%2Bx%2B1%7D%2C%7Bx%2C-10%2C10%7D%5D

Answer 21

Honestly just graphing this is the best way to do this lol.

Answer 22

That's why Descartes invented graphing in the first place, to make stuff like this easier. And then we got computers, lol.

Answer 23

i suggest that we study \[y=1+\frac{2x}{x^2+x+1} ~~x\in \mathbb{R}\]

Answer 24

minus not plus mistake haha

Answer 25

\[y=1-\frac{2x}{x^2+x+1}\]

Answer 26

hmm need to find upper and lower bound
m<y<M
some how i think that finding the range of x for 2x/x^2+x+1 might be our way!!!

Answer 27

I'm just throwing stuff as i go lol

Answer 28

x^2+x+1=0
has no real solution yes? so this parabola is >0

Answer 29

We can factor it and for just simplicity of writing I'll make this: \[\omega = e^{i 2 \pi / 3}\] \[\frac{x^2-x+1}{x^2+x+1}=\frac{(x+\omega)(x+\omega^2)}{(x-\omega)(x-\omega^2)}\]

Answer 30

y1=2x/x^2+x+1 can we find the range for this going by what i just did?

Answer 31

let \(x=\tan z\) with \(z \in (-\frac{\pi}{2}, \frac{\pi}{2})\) you will get\[f(\tan z)=\frac{\tan^2 z -\tan z+1}{\tan^2 z +\tan z+1}=\frac{4}{2+\sin (2z)}-1\]

Answer 32

hmm interesting @mukushla

Answer 33

i feel that my way can be a lot easier haha though don't you think

Answer 34

just need to find the range for 2x/(x^2+x+1)?

Answer 35

yeah, making a perfect square in denom, that's a lot better than mine

Answer 36

need pen and paper!

Answer 37

i like that tan substitution looks something creative

Answer 38

let me continue with yours\[f(x)=1-\frac{2x}{x^2+x+1}=1-\frac{2}{x+\frac{1}{x}+1}\]actually there is no need to make perfect square

Answer 39

we just need to note that\[\left|x+\frac{1}{x} \right| \ge 2 \]

Answer 40

hmm seems you a different approach

Answer 41

i was trying to use parabola and eyeball the range

Answer 42

how is |x+1/x|>=2

Answer 43

first time i see that :)

Answer 44

AM-GM inequality\[x+\frac{1}{x} \ge 2\sqrt{x.\frac{1}{x}}\]

Answer 45

if \(x<0\) it will be\[x+\frac{1}{x} \le -2\]

Answer 46

hmm i see

Answer 47

equality occurs when x=-1 and x=1 for two cases

Answer 48

so for x=-1
we have x+1/x<-2==> x+1/x+1<-1
then y<3

Answer 49

for x=1
by the same reasoning
1/3<y

Answer 50

excellent @mukushla

Answer 51

That's a nice trick with that AM-GM inequality.

Answer 52

beautiful indeed :)

Answer 53

@mathmath333 ty, actually it is a nice trick to use trig sub for finding range, you may try this one\[f(x)=\sqrt{x-2}+\sqrt{4-x}\]

Answer 54

thats incorrect

Answer 55

hmm if one realize they can use trig sub
the problem becomes really easy
since sin and cos are easily bounded

Answer 56

that's right

Answer 57

the range of x here is \(2<x<4\)

Answer 58

when you wanna do a trig sub, find the domain, it will give you the clue for the form of sub

Answer 59

well domain above is R
how can that help

Answer 60

tan covers R

Answer 61

but tan has pi/2 singularity?

Answer 62

does not matter ?

Answer 63

no we will treat bounds in limit, so it doesn't matter

Answer 64

now, what about the second function?

Answer 65

note that\[2 \le x \le4\]is a restricted domain

Answer 66

yeah just about to say that
math did it already

Answer 67

so what sub is that telling us to use?

Answer 68

sin or cos
?

Answer 69

hold on don't tell me yet

Answer 70

both of then can be

Answer 71

you know why?

Answer 72

hmm no really
because they are both have same bounds -1 and 1

Answer 73

yes and because\[ -1\le x-3 \le 1\]

Answer 74

oh i see

Answer 75

They're both the same function just shifted argument by 90 degrees.

\(\cos x + 3\) will work just as well as \(\sin x + 3\)

Answer 76

quite right

Answer 77

so in that we put x-3=sinz
yes?

Answer 78

damn i feel weak in trig lol :)

Answer 79

oh yeah, \(3+\cos z\) will be better maybe

Answer 80

hmm you said they all work fine No?
okay lets go with x=3+cosz

Answer 81

well derive with cos will be easier

Answer 82

so \[y=\sqrt{\cos z+1}+\sqrt{1-\cos z}\]

Answer 83

so \[0\leq y \leq 2\sqrt{2}\]

Answer 84

oh no you can't just plug in there values, your trig function is not alone, you'll need more simplification

Answer 85

if we simplify we just get radical 2?

Answer 86

if i didn't make an error that is

Answer 87

\[f=\sqrt{2} \left( \left|\sin \frac{z}{2} \right|+\left|\cos \frac{z}{2} \right| \right)\]

Answer 88

cosz=2cos^2x/2-1
oh i see i threw the square when i did my simplification haha

Answer 89

i didn't i mean*

Answer 90

ok, good discussion, tnx guys

Answer 91

well isn't that the same result above?
0<y<2root2

Answer 92

thanks to you learned bunch of stuff here :)

Answer 93

this one i never remember
\[\frac{a+b}{2}\leq \sqrt{ab}\]