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

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CliffSedgeBest ResponseYou've already chosen the best response.1
Do you know how to find derivatives?
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
You can try it that way, but of course that will involve solving a 7th degree polynomial...
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
ahh,,i'd have never thought of that! ;)
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
I also recommend graphing it as you go to help narrow down where it might be.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
well these solutions seem fine but are quite tedious,,dont you think ?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
and afterall,,its no olympiad problem, just a 12th standard problem..so has so be a quick solution.,.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Yes, but here's some good news, the minimum occurs as integer values of (x,y).
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
^ LOL, yeah, my TI84 already solved this for me, but I like doing these things by hand too.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
It's actually a pretty quick one to find using rational root theorem and synthetic division.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
I'm also checking to see if the first derivative is factorable.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Well, here's what I did: \[\large f'(x)=8x^748x^5+76x^336x^2+28x8\] 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.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Followed by second derivative test, a little pointplotting to verify, and yeah, not too bad.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
i'd still look forward to a shorter solution ! ..though i completely appreciate your method..
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Is this for a calculus course or algebra/precalc?
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
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.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
am not doing any school work sir..am preparing for ISI .. the question is in its practice paper.. treat is a 12+ competetice exam..
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
You want the global or absolute minimum, correct?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.0
\[(x^8 8x^6 + 19x^4 12x^3 +14x^2 8x +8)+1\]
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
x^8 8x^6 + 19x^4 12x^3 +14x^2 8x +9 x^8+19x^4(8x^6 +12x^3) +(14x^28x)+9 (x^419/2)^2 +14 (x^24)^28(x^3+6)^217.25
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
So, minimum when x^24=0 Thus, x=+2
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.0
Which gives 1543
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.0
@sauravshakya, there are errors in your factoring above.
 one year ago

estudierBest ResponseYou've already chosen the best response.0
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.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
ans. is 1 i tired factoring as @mukushla said.. i got somewhere but not enough progress..
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
Did you try the first and second derivative tests like I did? It was pretty easy to track down the solution that way.
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
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..
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
calcminimum function on a graphing calculator? I don't know anything shorter or sweeter than that. ;)
 one year ago

estudierBest ResponseYou've already chosen the best response.0
I doubt there is anything faster by hand than what Cliffsedge has given already....
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
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??
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
not confirmed,,i mean obtained*
 one year ago

estudierBest ResponseYou've already chosen the best response.0
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....
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.1
I agree, finding the root of f(x)1 seems like too much of a lucky guess.
 one year ago
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