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 
(ab)^2>=0 (where a and b are real numbers, assuming) a^22ab+b^2=a^2+2ab+b^24ab>=0 a^2+2ab+b^2=(a+b)^2>=4ab a+b>=2âˆšab, 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)>=2âˆš(4^x)(4^x)=2 (2^x)+(2^x)>=2âˆš(2^x)(2^x)=2 we have y=24k. since we're trying to find the number k (k<2), subtitute 2 for y. 2=24k 2=4k2 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>=2âˆšab, 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 + (2^x  k)²  2k² instead. 

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