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July 31st, 2007, 04:51 PM   #1
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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.
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July 31st, 2007, 11:19 PM   #2
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(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.
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August 4th, 2007, 10:06 PM   #3
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Unfortunately, johnny, your reasoning is unsound.

Try using y = (2^x - k) + (2^-x - k) - 2k instead.
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