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 April 21st, 2012, 11:54 PM #1 Newbie   Joined: Apr 2012 Posts: 18 Thanks: 0 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
 April 22nd, 2012, 12:15 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: minimum value of function If im not misunderstanding you, then the answer is d) , 9
 April 22nd, 2012, 12:22 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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. $f(1)=5$
April 22nd, 2012, 12:25 AM   #4
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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

 April 22nd, 2012, 12:39 AM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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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