Find the minimum value. 
Find  the minimum value of $4^x + 4^{1-x}$  , $x\in\mathbb{R}$.

In this I used the property that $a + \frac{1}{a}\geq 2$.
So I begin with 
$$ 4^x + \left(\frac{1}{4}\right)^x + 3\left(\frac{1}{4}\right)^x \geq 2 + 3\left(\frac{1}{4}\right)^x$$
So I think the minimum value be between 2 to 3.
But the answer is 4.
Thanks in advance.
 A: $$4^{x} + 4^{1-x} = (\sqrt{4^x} - \frac{2}{\sqrt{4^x}})^2 + 4 \ge 4$$
with equality occuring when $\sqrt{4^x} = \frac{2}{\sqrt{4^x}}$ and so $x = \frac{1}{2}$.
A: Your approach was a good one, except for the lack of "$x$, $-x$" symmetry.
But that is not hard to fix. Since $1/2$ is the midway point between $0$ and $1$,  it seems natural to look instead at $4^{x-1/2}+4^{1/2-x}$. This is a close relative of our expression, since
$$4^x+4^{1-x}=4^{1/2}(4^{x-1/2}+4^{1/2-x}).$$
Now it's over. By the result you quoted, the expression $4^{x-1/2}+4^{1/2-x}$ reaches a minimum of $2$ when $x=1/2$. So the minimum value of our original expression is $(4^{1/2})(2)$, which is $4$.
An alternate way of viewing the matter is that we made the symmetrizing change of variable $u=x-1/2$.  Then $4^x=4^{1/2}4^u$ and $4^{1-x}=4^{1/2}4^{-u}$. The procedure is quite analogous to what happens when we complete the square in $x^2-x$. There, we can let $u=x-1/2$, and the parabola $y=x^2-x$ becomes the nicer parabola $y=u^2-\frac{1}{4}$, from which the minimum can be read off. 
