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shubhamsrg

  • 2 years ago

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

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  1. CliffSedge
    • 2 years ago
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    Do you know how to find derivatives?

  2. shubhamsrg
    • 2 years ago
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    yes..i do..

  3. CliffSedge
    • 2 years ago
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    You can try it that way, but of course that will involve solving a 7th degree polynomial...

  4. shubhamsrg
    • 2 years ago
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    ahh,,i'd have never thought of that! ;)

  5. CliffSedge
    • 2 years ago
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    I also recommend graphing it as you go to help narrow down where it might be.

  6. shubhamsrg
    • 2 years ago
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    well these solutions seem fine but are quite tedious,,dont you think ?

  7. shubhamsrg
    • 2 years ago
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    and afterall,,its no olympiad problem, just a 12th standard problem..so has so be a quick solution.,.

  8. CliffSedge
    • 2 years ago
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    Yes, but here's some good news, the minimum occurs as integer values of (x,y).

  9. CliffSedge
    • 2 years ago
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    ^ LOL, yeah, my TI-84 already solved this for me, but I like doing these things by hand too.

  10. CliffSedge
    • 2 years ago
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    It's actually a pretty quick one to find using rational root theorem and synthetic division.

  11. CliffSedge
    • 2 years ago
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    I'm also checking to see if the first derivative is factorable.

  12. CliffSedge
    • 2 years ago
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    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.

  13. CliffSedge
    • 2 years ago
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    Followed by second derivative test, a little point-plotting to verify, and yeah, not too bad.

  14. shubhamsrg
    • 2 years ago
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    i'd still look forward to a shorter solution ! ..though i completely appreciate your method..

  15. CliffSedge
    • 2 years ago
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    Is this for a calculus course or algebra/pre-calc?

  16. CliffSedge
    • 2 years ago
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    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.

  17. shubhamsrg
    • 2 years ago
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    am not doing any school work sir..am preparing for ISI .. the question is in its practice paper.. treat is a 12+ competetice exam..

  18. shubhamsrg
    • 2 years ago
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    competitive*

  19. calculusfunctions
    • 2 years ago
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    You want the global or absolute minimum, correct?

  20. mukushla
    • 2 years ago
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    try to factor it

  21. mukushla
    • 2 years ago
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    \[(x^8 -8x^6 + 19x^4 -12x^3 +14x^2 -8x +8)+1\]

  22. sauravshakya
    • 2 years ago
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    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

  23. sauravshakya
    • 2 years ago
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    So, minimum when x^2-4=0 Thus, x=+-2

  24. sauravshakya
    • 2 years ago
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    Which gives -1543

  25. calculusfunctions
    • 2 years ago
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    @sauravshakya, there are errors in your factoring above.

  26. estudier
    • 2 years ago
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    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.

  27. shubhamsrg
    • 2 years ago
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    ans. is 1 i tired factoring as @mukushla said.. i got somewhere but not enough progress..

  28. CliffSedge
    • 2 years ago
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    Did you try the first and second derivative tests like I did? It was pretty easy to track down the solution that way.

  29. shubhamsrg
    • 2 years ago
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    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..

  30. CliffSedge
    • 2 years ago
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    calc-minimum function on a graphing calculator? I don't know anything shorter or sweeter than that. ;-)

  31. estudier
    • 2 years ago
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    I doubt there is anything faster by hand than what Cliffsedge has given already....

  32. shubhamsrg
    • 2 years ago
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    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??

  33. shubhamsrg
    • 2 years ago
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    not confirmed,,i mean obtained*

  34. estudier
    • 2 years ago
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    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....

  35. CliffSedge
    • 2 years ago
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    I agree, finding the root of f(x)-1 seems like too much of a lucky guess.

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