How is the answer local min. : 2^(-2/3) + 2^(1/3) and is CU on the interval (-∞,∞)

Find the local maximum and minimum values, the intervals of concavity and the inflection points for:

f(x)= e^(2x) + e^(-x).

f(x) = e^(2x) + e^(-x)

f ' (x) = 2e^(2x) - e^(-x)

f" = 4e^(2x) + e^(-x)
second question first:
4e^(2x) + e^(-x) is always positive, since an exponential function is always positive (with a positive leading coefficient)

f ' (x) = 0 when 2e^(2x) = e^(-x)
2e^(2x) = 1 / (e^x)
e^(3x) = 1/2
3x = ln (1/2)
x = ln (1/2) / 3
x = ln (1/2)^(1/3) = ln (2^(-1/3))

f(ln (2^(-1/3)) = e^(2 ln 2^(-1/3)) + e^(-ln 2^(-1/3))
= e^(ln (2^(-2/3)) + e^(ln 2^(1/3))

and since e^(ln u) = u, then f(ln (2^(-1/3)) = 2^(-2/3) + 2^(1/3)

f " > 0, so this is a local min

coordinates: (ln 2^(-2/3) , 2^(-2/3) + 2^(1/3))

What is wrong with the answers?

f ' (x) = 2e^(2x) - e^(-x) ==> f '(x) = 0 if x = -ln(2)/3.

When x = -ln(2)/3 = ln(2^(-1/3))

f(ln(2^(-1/3)) = 2^(-2/3) + 2^(1/3).

Moreover,

f ''(x) = 4e^(2x) + e^(-x) > 0 for all real x.

So the graph is everywhere "concave up".

local min

{3/2^(2/3), {x = -(Log[2]/3)}}