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

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CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1Do you know how to find derivatives?

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

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0ahh,,i'd have never thought of that! ;)

CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1I also recommend graphing it as you go to help narrow down where it might be.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0well these solutions seem fine but are quite tedious,,dont you think ?

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

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

CliffSedge
 3 years ago
Best 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.

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

CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1I'm also checking to see if the first derivative is factorable.

CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1Well, 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.

CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1Followed by second derivative test, a little pointplotting to verify, and yeah, not too bad.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0i'd still look forward to a shorter solution ! ..though i completely appreciate your method..

CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1Is this for a calculus course or algebra/precalc?

CliffSedge
 3 years ago
Best ResponseYou've already chosen the best response.1Have 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
 3 years ago
Best ResponseYou've already chosen the best response.0am not doing any school work sir..am preparing for ISI .. the question is in its practice paper.. treat is a 12+ competetice exam..

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.0You want the global or absolute minimum, correct?

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

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0x^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

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0So, minimum when x^24=0 Thus, x=+2

calculusfunctions
 3 years ago
Best ResponseYou've already chosen the best response.0@sauravshakya, there are errors in your factoring above.

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0http://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
 3 years ago
Best ResponseYou've already chosen the best response.0ans. is 1 i tired factoring as @mukushla said.. i got somewhere but not enough progress..

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

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0indeed,,your solution yielded the correct ans,,i.e. minima at x=2 .. and i appreciate that.. but still looking for shorter and sweeter solution..

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

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0I doubt there is anything faster by hand than what Cliffsedge has given already....

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.0not confirmed,,i mean obtained*

estudier
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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....

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