Question

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

Answer 1

`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.

[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`

Answer 2

`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.!