Finding the Range of the given function

[Kaushik]

Homework Statement

Find the range of $f(x) = 6^x + 3^x + 6^{-x} + 3^{-x} + 2$

Relevant Equations

AM >= GM

$f(x) = 6^x + 3^x + 6^{-x} + 3^{-x} + 2$

But, $AM >= GM$

So,

$f(x) >= 5 * 2 ^ {\frac{1}{5}}$

But this is not the case. According to the graph, it is $f(x) >= 6$.

If I do the same thing without considering the constant '2' then I am getting the answer.

let $g(x) = 6^x + 3^x + 6^{-x} + 3^{-x}$ and $f(x) = g(x) + 2$

Using $AM>=GM$,

$g(x) >= 4$

Hence, $f(x) >= 6$

Why is this the case? Why is the latter approach working but no the former?

 

[Mark44]

Homework Statement:: Find the range of $f(x) = 6^x + 3^x + 6^{-x} + 3^{-x} + 2$
Relevant Equations:: AM >= GM

$f(x) = 6^x + 3^x + 6^{-x} + 3^{-x} + 2$

But, $AM >= GM$
So,
$f(x) >= 5 * 2 ^ {\frac{1}{5}}$

I don't see what AM >= GM has to do with this problem, or how the inequality you show is related to this problem. Certainly $f(x) \ge 5 \cdot 2^{1/5}$, but that's a very rough lower bound.

But this is not the case. According to the graph, it is $f(x) >= 6$.

View attachment 258864

If I do the same thing without considering the constant '2' then I am getting the answer.

let $g(x) = 6^x + 3^x + 6^{-x} + 3^{-x}$ and $f(x) = g(x) + 2$

Using $AM>=GM$,

$g(x) >= 4$

Hence, $f(x) >= 6$

Why is this the case? Why is the latter approach working but no the former?

$6^x + 6^{-x}$ has a minimum value of 2 for x = 0. Likewise, $3^x + 3^{-x}$ also has a minimum value of 2 at x = 0. Add these pairs together + 2, and you get a minimum value of 6.