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 July 31st, 2007, 04:51 PM #1 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 minimum values if the absolute minimum value of graph y=(4^x)+(4^-x)-2k((2^x)+(2^-x)) is -2, then find the real number value for k, where k<2. show work algebraically - don't use calculus, and don't use tools such as graphing calculators. July 31st, 2007, 11:19 PM #2 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 (a-b)^2>=0 (where a and b are real numbers, assuming) a^2-2ab+b^2=a^2+2ab+b^2-4ab>=0 a^2+2ab+b^2=(a+b)^2>=4ab a+b>=2√ab, where a>=0 and b>=0 use this general equation for (4^x)+(4^-x) and (2^x)+(2^-x), because we want the absolute minimum value of y. (4^x)+(4^-x)>=2√(4^x)(4^-x)=2 (2^x)+(2^-x)>=2√(2^x)(2^-x)=2 we have y=2-4k. since we're trying to find the number k (k<2), subtitute -2 for y. -2=2-4k 2=4k-2 4k=4 k=1... which is the final answer to this problem, and solved algebraically without the use of calculus and/or graphing calculator. if anyone of you got different solutions, please post them. btw.. keep in mind that a+b>=2√ab, where a>=0 and b>=0. it could be very useful. August 4th, 2007, 10:06 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2201 Unfortunately, johnny, your reasoning is unsound. Try using y = (2^x - k)� + (2^-x - k)� - 2k� instead. Tags minimum, values Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Telu Calculus 4 August 2nd, 2013 10:00 PM space55 Calculus 0 October 10th, 2010 03:23 PM winnerhere Elementary Math 3 September 1st, 2010 03:36 AM dk1702 Calculus 4 May 11th, 2010 05:58 PM K Sengupta Number Theory 2 March 5th, 2009 06:12 AM

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