Find range

Since $4^{x^2} + 4^{(x-1)^2} > 0$ and continuous for all $x$ , $y$ is continuous and $y> 0$ for all $x$ . Therefore, it has a minimum value. Minimum $y$ occurs when $4^{x^2} + 4^{(x-1)^2}$ is minimum. By AM-GM inequality , we have:

$\begin{aligned} 4^{x^2} + 4^{(x-1)^2} & \ge 2 \sqrt{4^{x^2+(x-1)^2}} = 2 \cdot 2^{x^2+(x-1)^2} \end{aligned}$

Equality occurs when $4^{x^2} = 4^{(x-1)^2}$ or $x^2 = (x-1)^2$ or $x = \frac 12$ , then $y = \log_2 (2\sqrt 3) = \frac 32 = \boxed {0.5}$ .