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.!