Finding the range of $f(x)=\log_2\left(4^{x^2}+4^{(x-1)^2}\right)$ 
If the range of the function $f(x)=\log_2\left(4^{x^2}+4^{(x-1)^2}\right)$ is $\left(\frac pq,\infty\right)$, where $p, q$ are in their lowest form then find $(p+q), (p-q)$

Looks like the domain is $\mathbb R$.
If $x^2=0$, $(x-1)^2=1\implies f(0)=\log_2\left(1+4\right)=\log_25$
If $(x-1)^2=0, x^2=1\implies f(1)=\log_25$
So, I think $p=\log_25, q=1$
So, $p+q=\log_25+1, p-q=\log_25-1$
But the options are $1/3/4/5/10$.
I tried converting the base to $10$ and writing $5$ as $10/2$ but in vain.
 A: By Jensen for the convex function $f(x)=4^{x^2}$ we obtain:
$$4^{x^2}+4^{(x-1)^2}\geq2\cdot4^{\left(\frac{x+1-x}{2}\right)^2}=2\sqrt2.$$
Can you end it now?
A: It seems that there is an error in the statement of the problem, since the range of $f$ is of the form $\left[\frac pq,\infty\right)$.
Anyway,$$f'(x)=4\left(x-\frac1{1+4^{2x-1}}\right).$$Since $x\mapsto x$ is strictly increasing and $x\mapsto\frac1{1+4^{2x-1}}$ is strictly decreasing, $f'(x)$ has at most one zero. But it's clear that $f'\left(\frac12\right)=0$. It follows from what I wrote that $f'(x)<0$ if $x<\frac12$ and that $f'(x)>0$ if $x>\frac12$. So, the minimum of $f$ is attained at $\frac12$. Since $f\left(\frac12\right)=\frac32$ and since $\lim_{x\to\pm\infty}f(x)=\infty$, the range of $f$ is $\left[\frac32,\infty\right)$. So, $p=3$, and $q=2$.
A: Note that $4^{x^2}+4^{(x-1)^2}$ is strictly positive on $\mathbb R$, so taking $\log_2$ of it is well defined everywhere on $\mathbb R$. Moreover it is a composition of continuously differentiable ($C^1$) functions hence $C^1$ on $\mathbb R$. As $x\to \infty$ and $x\to-\infty$, $4^{x^2}+4^{(x-1)^2}$ goes to $\infty$, so $f$ goes to $\infty$ as $x\to \pm \infty$. With this information plus the fact that $f$ is $C^1$ on $\mathbb R$, we know that the (global) minimum of $f$ is attained at some $x_0$ and $x_0$ satisfies $f'(x_0)= 0$. You can compute this yourself and find $x_0$. Lastly, continuity means that $f$ satisfies the intermediate value property, so indeed $f$ hits EVERY value $\geq f(x_0)$, meaning the range is $[f(x_0),\infty)$.
A: You have
$$\begin{align}4^{x^2}+4^{{(x-1)}^2}
&≥2\sqrt{4^{x^2+(x-1)^2}}\\
&=2^{{x^2+(x-1)^2}+1}\\
&≥2^{2\left(x-\frac 12\right)^2+\frac 32}\\
&≥2^{\frac 32}\end{align}$$
$$\log_2\left(4^{x^2}+4^{(x-1)^2}\right)≥\log_2{2^{\frac 32}}=\frac 32.$$
A: We have that by $x=u+\frac12$ and AM-GM
$$g(x)=4^{x^2}+4^{(x-1)^2}=4^{\left(u+\frac12\right)^2}+4^{\left(u-\frac12\right)^2}=$$
$$=h(u)=4^{\left(u^2+\frac14\right)}\left(4^u+4^{-u}\right)\ge 2\sqrt 2$$
with $h(u):\mathbb R \to [2\sqrt 2, \infty)$ even and increasing and since also $\log_2$ function is ingreasing we can conclude that $$f(x)\ge \log_2 2\sqrt 2=\frac32$$
