The range of function f(x ) = 4^x + 2^x + 4^-x + 2^-x + 3 is

$4^x + 2^x + 4^{-x} + 2^{-x} + 3=t^2+\frac{1}{t^2}+t+\frac{1}{t}+3$, where $t=2^x>0, ∀\,x∈R$

$⇒\left(t+\frac{1}{t}\right)^2+\left(t+\frac{1}{t}\right)+1$

since $t+\frac{1}{t}>2, ∀\,t>0$  $∴f(x)>2^2+2+1=7$  ∴ Range is (7, ∞)