shubhamsrg
  • shubhamsrg
the minimum value of x^8 -8x^6 + 19x^4 -12x^3 +14x^2 -8x +9 = ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Do you know how to find derivatives?
shubhamsrg
  • shubhamsrg
yes..i do..
anonymous
  • anonymous
You can try it that way, but of course that will involve solving a 7th degree polynomial...

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shubhamsrg
  • shubhamsrg
ahh,,i'd have never thought of that! ;)
anonymous
  • anonymous
I also recommend graphing it as you go to help narrow down where it might be.
shubhamsrg
  • shubhamsrg
well these solutions seem fine but are quite tedious,,dont you think ?
shubhamsrg
  • shubhamsrg
and afterall,,its no olympiad problem, just a 12th standard problem..so has so be a quick solution.,.
anonymous
  • anonymous
Yes, but here's some good news, the minimum occurs as integer values of (x,y).
anonymous
  • anonymous
^ LOL, yeah, my TI-84 already solved this for me, but I like doing these things by hand too.
anonymous
  • anonymous
It's actually a pretty quick one to find using rational root theorem and synthetic division.
anonymous
  • anonymous
I'm also checking to see if the first derivative is factorable.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Followed by second derivative test, a little point-plotting to verify, and yeah, not too bad.
shubhamsrg
  • shubhamsrg
i'd still look forward to a shorter solution ! ..though i completely appreciate your method..
anonymous
  • anonymous
Is this for a calculus course or algebra/pre-calc?
anonymous
  • anonymous
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.
shubhamsrg
  • shubhamsrg
am not doing any school work sir..am preparing for ISI .. the question is in its practice paper.. treat is a 12+ competetice exam..
shubhamsrg
  • shubhamsrg
competitive*
calculusfunctions
  • calculusfunctions
You want the global or absolute minimum, correct?
anonymous
  • anonymous
try to factor it
anonymous
  • anonymous
\[(x^8 -8x^6 + 19x^4 -12x^3 +14x^2 -8x +8)+1\]
anonymous
  • anonymous
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
anonymous
  • anonymous
So, minimum when x^2-4=0 Thus, x=+-2
anonymous
  • anonymous
Which gives -1543
calculusfunctions
  • calculusfunctions
@sauravshakya, there are errors in your factoring above.
anonymous
  • anonymous
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.
shubhamsrg
  • shubhamsrg
ans. is 1 i tired factoring as @mukushla said.. i got somewhere but not enough progress..
anonymous
  • anonymous
Did you try the first and second derivative tests like I did? It was pretty easy to track down the solution that way.
shubhamsrg
  • shubhamsrg
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..
anonymous
  • anonymous
calc-minimum function on a graphing calculator? I don't know anything shorter or sweeter than that. ;-)
anonymous
  • anonymous
I doubt there is anything faster by hand than what Cliffsedge has given already....
shubhamsrg
  • shubhamsrg
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??
shubhamsrg
  • shubhamsrg
not confirmed,,i mean obtained*
anonymous
  • anonymous
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....
anonymous
  • anonymous
I agree, finding the root of f(x)-1 seems like too much of a lucky guess.

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