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 amateurmathlover April 21st, 2012 11:54 PM

minimum value of function

The minimum value of f(x)=x^8+x^6-x^4-2x^3-x^2-2x+9 is:
a 5
b)1
c)0
d)9

 mathbalarka April 22nd, 2012 12:15 AM

Re: minimum value of function

If im not misunderstanding you, then the answer is d) , 9

 MarkFL April 22nd, 2012 12:22 AM

Re: minimum value of function

The derivative of f equated to zero has only 1 real root, which is $x=1$.

$f''(1)>0$ meaning there is a minimum there and since f has an even order, we know this is a global minimum.

$f(1)=5$

 amateurmathlover April 22nd, 2012 12:25 AM

Re: minimum value of function

Quote:
 Originally Posted by MarkFL The derivative of f equated to zero has only 1 real root, which is $x=1$. $f''(1)>0$ meaning there is a minimum there and since f has an even order, we know this is a global minimum. $f(1)=5$
Thanks. The problem I was facing was with finding the roots of the first derivative. Can u please explain how u have found that? Thanks again :D

 MarkFL April 22nd, 2012 12:39 AM

Re: minimum value of function

I used the rational roots theorem, and checked, using polynomial division to find:

$8x^7+6x^5-4x^2-6x^2-2x-2=2(x-1)\(4x^6+4x^5+7x^4+7x^3+7x^2+2x+1\)$

I then found numerically that the remaining 6th order factor has a global minimum greater than zero, hence it has no real roots.

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