minimum value of 5^x + 3^x + e^x + 5^-x + 3^-x + e^-x ?
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"What is a supersaturated solution?"
#f(x) = 5^x+3^x+e^x+5^-x+3^-x+e^-x#
#= (5^x+1/5^x) + (3^x+1/3^x) + (e^x+1/e^x)#
Now each of the three terms above has a minimum value of #2# at #x=0# as shown graphically below.
![enter image source here]()
[Red #= (5^x+1/5^x)# Blue #= (3^x+1/3^x)# Green #= (e^x+1/e^x)#]
Hence #f(x)_min = 2+2+2 =6#
Prerequisite : AM-GM Property :
If #a,b# are positive reals, then their #AM ge GM.#
#AM=(a+b)/2, GM=sqrt(ab)#.
The given expression (exp.) can be rewritten as,
#"The exp."=(5^x+5^-x)+(3^x+3^-x)+(e^x+e^-x)#.
Applying the AM-GM Property to positive reals, #5^x and 5^-x#,
we have, #(5^x+5^-x)/2 ge sqrt(5^x*5^-x), i.e., #
#(5^x+5^-x)/2 ge 1, or, (5^x+5^-x) ge2#.
Likewise, #(3^x+3^-x) ge 2, and, (e^x+e^-x) ge 2#.
Adding these, we get,
#(5^x+5^-x)+(3^x+3^-x)+(e^x+e^-x) ge 6#, giving the
minimum value of the exp., #6#, as Respected Alan N.
has readily derived!
Enjoy Maths.!
