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the minimum value of x^8 -8x^6 + 19x^4 -12x^3 +14x^2 -8x +9 = ?

Mathematics
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Do you know how to find derivatives?
yes..i do..
You can try it that way, but of course that will involve solving a 7th degree polynomial...

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Other answers:

ahh,,i'd have never thought of that! ;)
I also recommend graphing it as you go to help narrow down where it might be.
well these solutions seem fine but are quite tedious,,dont you think ?
and afterall,,its no olympiad problem, just a 12th standard problem..so has so be a quick solution.,.
Yes, but here's some good news, the minimum occurs as integer values of (x,y).
^ LOL, yeah, my TI-84 already solved this for me, but I like doing these things by hand too.
It's actually a pretty quick one to find using rational root theorem and synthetic division.
I'm also checking to see if the first derivative is factorable.
Well, here's what I did: \[\large f'(x)=8x^7-48x^5+76x^3-36x^2+28x-8\] Factor out a 4 and set equal to zero. \[\large 2x^7−12x^5+19x^3−9x^2+7x−2=0\] Possible rational roots ε ± {1, 2, 1/2} Tested 1, didn't work, tested 2, found it.
Followed by second derivative test, a little point-plotting to verify, and yeah, not too bad.
i'd still look forward to a shorter solution ! ..though i completely appreciate your method..
Is this for a calculus course or algebra/pre-calc?
Have you seen similar examples from your teacher or in your book? What tools are you expected to use? Are you required to do it by hand, or are graphing calculators allowed? All these considerations will determine the best way to do it.
am not doing any school work sir..am preparing for ISI .. the question is in its practice paper.. treat is a 12+ competetice exam..
competitive*
You want the global or absolute minimum, correct?
try to factor it
\[(x^8 -8x^6 + 19x^4 -12x^3 +14x^2 -8x +8)+1\]
x^8 -8x^6 + 19x^4 -12x^3 +14x^2 -8x +9 x^8+19x^4-(8x^6 +12x^3) +(14x^2-8x)+9 (x^4-19/2)^2 +14 (x^2-4)^2-8(x^3+6)^2-17.25
So, minimum when x^2-4=0 Thus, x=+-2
Which gives -1543
@sauravshakya, there are errors in your factoring above.
http://www.willamette.edu/~mjaneba/help/Polynomial%20root%20finding.html Frankly, the purpose of this sort of thing (ie large degree polynomial by hand) can only be just the practice of math skills and to help remember all those bits and pieces about polynomials.... And, since it has been set as a question, you can assume there is a (relatively) easy solution of some sort.
ans. is 1 i tired factoring as @mukushla said.. i got somewhere but not enough progress..
Did you try the first and second derivative tests like I did? It was pretty easy to track down the solution that way.
indeed,,your solution yielded the correct ans,,i.e. minima at x=2 .. and i appreciate that.. but still looking for shorter and sweeter solution..
calc-minimum function on a graphing calculator? I don't know anything shorter or sweeter than that. ;-)
I doubt there is anything faster by hand than what Cliffsedge has given already....
i was able to check that(following @mukushla 's work) x=2 was a root of f(x) - 1 which can be confirmed from rational root theorem,, will that be satisfactory??
not confirmed,,i mean obtained*
Well, that's about as good as making an inspired guess that there is a root of 2. If you want to go down that route, then say to yourself well, the x^8 obviously takes over for x greater than plus or minus 3, so test 0,1,2 and voila! you found it....
I agree, finding the root of f(x)-1 seems like too much of a lucky guess.

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