mathmath333
  • mathmath333
Find the minimum and maximum value of
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} \dfrac{x^2-x+1}{x^2+x+1},\ x\in \mathbb{R}\hspace{.33em}\\~\\ \end{align}}\)
anonymous
  • anonymous
It's hard to explain...but I will see if I can show the work.
anonymous
  • anonymous
are you talking local or absolute maxes and mins?

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mathmath333
  • mathmath333
i would also prefer a way without calculus ( idk calculus)
anonymous
  • anonymous
Do you need to show your work for this problem?
mathmath333
  • mathmath333
not necessary i should understand the idea
anonymous
  • anonymous
Based on what I know algebraically , I got a minimum of (1, 1/3) and a maximum of (-1,3).
mathmath333
  • mathmath333
u calculated the co-ordinates where as it should be the numerical value
anonymous
  • anonymous
Ok. Let me try a different way.
anonymous
  • anonymous
Ok. I got the minimum is 1/3, and the maximum is -3.
anonymous
  • anonymous
oops +3. not neg
mathmath333
  • mathmath333
how did u calculate
anonymous
  • 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
  • anonymous
Same for the denominator, except it is when x=1
anonymous
  • 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
  • xapproachesinfinity
eh i like your questions lol
xapproachesinfinity
  • xapproachesinfinity
make me struggle :)
mathmath333
  • mathmath333
i have tons of question but no time
xapproachesinfinity
  • xapproachesinfinity
i cheated haha min=1/3
mathmath333
  • 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
  • Empty
Honestly just graphing this is the best way to do this lol.
Empty
  • Empty
That's why Descartes invented graphing in the first place, to make stuff like this easier. And then we got computers, lol.
xapproachesinfinity
  • xapproachesinfinity
i suggest that we study \[y=1+\frac{2x}{x^2+x+1} ~~x\in \mathbb{R}\]
xapproachesinfinity
  • xapproachesinfinity
minus not plus mistake haha
xapproachesinfinity
  • xapproachesinfinity
\[y=1-\frac{2x}{x^2+x+1}\]
xapproachesinfinity
  • xapproachesinfinity
hmm need to find upper and lower bound m
xapproachesinfinity
  • xapproachesinfinity
I'm just throwing stuff as i go lol
xapproachesinfinity
  • xapproachesinfinity
x^2+x+1=0 has no real solution yes? so this parabola is >0
Empty
  • 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
  • xapproachesinfinity
y1=2x/x^2+x+1 can we find the range for this going by what i just did?
anonymous
  • 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
  • xapproachesinfinity
hmm interesting @mukushla
xapproachesinfinity
  • xapproachesinfinity
i feel that my way can be a lot easier haha though don't you think
xapproachesinfinity
  • xapproachesinfinity
just need to find the range for 2x/(x^2+x+1)?
anonymous
  • anonymous
yeah, making a perfect square in denom, that's a lot better than mine
xapproachesinfinity
  • xapproachesinfinity
need pen and paper!
mathmath333
  • mathmath333
i like that tan substitution looks something creative
anonymous
  • 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
  • anonymous
we just need to note that\[\left|x+\frac{1}{x} \right| \ge 2 \]
xapproachesinfinity
  • xapproachesinfinity
hmm seems you a different approach
xapproachesinfinity
  • xapproachesinfinity
i was trying to use parabola and eyeball the range
xapproachesinfinity
  • xapproachesinfinity
how is |x+1/x|>=2
xapproachesinfinity
  • xapproachesinfinity
first time i see that :)
anonymous
  • anonymous
AM-GM inequality\[x+\frac{1}{x} \ge 2\sqrt{x.\frac{1}{x}}\]
anonymous
  • anonymous
if \(x<0\) it will be\[x+\frac{1}{x} \le -2\]
xapproachesinfinity
  • xapproachesinfinity
hmm i see
anonymous
  • anonymous
equality occurs when x=-1 and x=1 for two cases
xapproachesinfinity
  • xapproachesinfinity
so for x=-1 we have x+1/x<-2==> x+1/x+1<-1 then y<3
xapproachesinfinity
  • xapproachesinfinity
for x=1 by the same reasoning 1/3
xapproachesinfinity
  • xapproachesinfinity
excellent @mukushla
Empty
  • Empty
That's a nice trick with that AM-GM inequality.
xapproachesinfinity
  • xapproachesinfinity
beautiful indeed :)
anonymous
  • 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
  • mathmath333
thats incorrect
xapproachesinfinity
  • xapproachesinfinity
hmm if one realize they can use trig sub the problem becomes really easy since sin and cos are easily bounded
anonymous
  • anonymous
that's right
mathmath333
  • mathmath333
the range of x here is \(2
anonymous
  • anonymous
when you wanna do a trig sub, find the domain, it will give you the clue for the form of sub
xapproachesinfinity
  • xapproachesinfinity
well domain above is R how can that help
anonymous
  • anonymous
tan covers R
xapproachesinfinity
  • xapproachesinfinity
but tan has pi/2 singularity?
xapproachesinfinity
  • xapproachesinfinity
does not matter ?
anonymous
  • anonymous
no we will treat bounds in limit, so it doesn't matter
anonymous
  • anonymous
now, what about the second function?
anonymous
  • anonymous
note that\[2 \le x \le4\]is a restricted domain
xapproachesinfinity
  • xapproachesinfinity
yeah just about to say that math did it already
xapproachesinfinity
  • xapproachesinfinity
so what sub is that telling us to use?
xapproachesinfinity
  • xapproachesinfinity
sin or cos ?
xapproachesinfinity
  • xapproachesinfinity
hold on don't tell me yet
anonymous
  • anonymous
both of then can be
anonymous
  • anonymous
you know why?
xapproachesinfinity
  • xapproachesinfinity
hmm no really because they are both have same bounds -1 and 1
anonymous
  • anonymous
yes and because\[ -1\le x-3 \le 1\]
xapproachesinfinity
  • xapproachesinfinity
oh i see
Empty
  • 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
  • anonymous
quite right
xapproachesinfinity
  • xapproachesinfinity
so in that we put x-3=sinz yes?
xapproachesinfinity
  • xapproachesinfinity
damn i feel weak in trig lol :)
anonymous
  • anonymous
oh yeah, \(3+\cos z\) will be better maybe
xapproachesinfinity
  • xapproachesinfinity
hmm you said they all work fine No? okay lets go with x=3+cosz
anonymous
  • anonymous
well derive with cos will be easier
xapproachesinfinity
  • xapproachesinfinity
so \[y=\sqrt{\cos z+1}+\sqrt{1-\cos z}\]
xapproachesinfinity
  • xapproachesinfinity
so \[0\leq y \leq 2\sqrt{2}\]
anonymous
  • anonymous
oh no you can't just plug in there values, your trig function is not alone, you'll need more simplification
xapproachesinfinity
  • xapproachesinfinity
if we simplify we just get radical 2?
xapproachesinfinity
  • xapproachesinfinity
if i didn't make an error that is
anonymous
  • anonymous
\[f=\sqrt{2} \left( \left|\sin \frac{z}{2} \right|+\left|\cos \frac{z}{2} \right| \right)\]
xapproachesinfinity
  • xapproachesinfinity
cosz=2cos^2x/2-1 oh i see i threw the square when i did my simplification haha
xapproachesinfinity
  • xapproachesinfinity
i didn't i mean*
anonymous
  • anonymous
ok, good discussion, tnx guys
xapproachesinfinity
  • xapproachesinfinity
well isn't that the same result above? 0
xapproachesinfinity
  • xapproachesinfinity
thanks to you learned bunch of stuff here :)
xapproachesinfinity
  • xapproachesinfinity
this one i never remember \[\frac{a+b}{2}\leq \sqrt{ab}\]

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