- mathmath333

Find the minimum and maximum value of

- jamiebookeater

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- mathmath333

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

- anonymous

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

- anonymous

are you talking local or absolute maxes and mins?

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## More answers

- mathmath333

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

- anonymous

Do you need to show your work for this problem?

- mathmath333

not necessary i should understand the idea

- anonymous

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

- mathmath333

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

- anonymous

Ok. Let me try a different way.

- anonymous

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

- anonymous

oops +3. not neg

- mathmath333

how did u calculate

- anonymous

\[\frac{x^2 - x + 1 }{ x^2 + x + 1 } = 3, when (x) = -1\]
I got this after simplifying to figure out what x = for the numerator

- anonymous

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

- anonymous

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

- xapproachesinfinity

eh i like your questions lol

- xapproachesinfinity

make me struggle :)

- mathmath333

i have tons of question but no time

- xapproachesinfinity

i cheated haha
min=1/3

- mathmath333

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

- Empty

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

- Empty

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

- xapproachesinfinity

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

- xapproachesinfinity

minus not plus mistake haha

- xapproachesinfinity

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

- xapproachesinfinity

hmm need to find upper and lower bound
m

- xapproachesinfinity

I'm just throwing stuff as i go lol

- xapproachesinfinity

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

- Empty

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

- xapproachesinfinity

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

- anonymous

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

- xapproachesinfinity

hmm interesting @mukushla

- xapproachesinfinity

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

- xapproachesinfinity

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

- anonymous

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

- xapproachesinfinity

need pen and paper!

- mathmath333

i like that tan substitution looks something creative

- anonymous

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

- anonymous

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

- xapproachesinfinity

hmm seems you a different approach

- xapproachesinfinity

i was trying to use parabola and eyeball the range

- xapproachesinfinity

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

- xapproachesinfinity

first time i see that :)

- anonymous

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

- anonymous

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

- xapproachesinfinity

hmm i see

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

for x=1
by the same reasoning
1/3

- xapproachesinfinity

excellent @mukushla

- Empty

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

- xapproachesinfinity

beautiful indeed :)

- anonymous

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

- mathmath333

thats incorrect

- xapproachesinfinity

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

- anonymous

that's right

- mathmath333

the range of x here is \(2

- anonymous

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

- xapproachesinfinity

well domain above is R
how can that help

- anonymous

tan covers R

- xapproachesinfinity

but tan has pi/2 singularity?

- xapproachesinfinity

does not matter ?

- anonymous

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

- anonymous

now, what about the second function?

- anonymous

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

- xapproachesinfinity

yeah just about to say that
math did it already

- xapproachesinfinity

so what sub is that telling us to use?

- xapproachesinfinity

sin or cos
?

- xapproachesinfinity

hold on don't tell me yet

- anonymous

both of then can be

- anonymous

you know why?

- xapproachesinfinity

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

- anonymous

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

- xapproachesinfinity

oh i see

- Empty

They're both the same function just shifted argument by 90 degrees.
\(\cos x + 3\) will work just as well as \(\sin x + 3\)

- anonymous

quite right

- xapproachesinfinity

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

- xapproachesinfinity

damn i feel weak in trig lol :)

- anonymous

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

- xapproachesinfinity

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

- anonymous

well derive with cos will be easier

- xapproachesinfinity

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

- xapproachesinfinity

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

- anonymous

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

- xapproachesinfinity

if we simplify we just get radical 2?

- xapproachesinfinity

if i didn't make an error that is

- anonymous

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

- xapproachesinfinity

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

- xapproachesinfinity

i didn't i mean*

- anonymous

ok, good discussion, tnx guys

- xapproachesinfinity

well isn't that the same result above?
0

- xapproachesinfinity

thanks to you learned bunch of stuff here :)

- xapproachesinfinity

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

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