The minimum value of 4^x+4^1-x, x∈R is

A.M ≥ G.M

$∴\frac{4^x+\frac{4}{4^x}}{2}≥\sqrt{4^x.\frac{4}{4^x}}$

$\Rightarrow 4^x+4^{1-x}\ge 4$

$\Rightarrow \text{Minimum value of } 4^x+4^{1-x} \text { is 4 }.$