minimum value of 5^x + 3^x + e^x + 5^-x + 3^-x + e^-x ?

2 Answers
Jun 12, 2018

#6#

Explanation:

#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#

Jun 12, 2018

# 6#.

Explanation:

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.!