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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.

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How did you get $4^{1-x} = (\frac{1}{4})^x + (\frac{3}{4})^x$? If you evaluate at $x=1$, you get $4$ on the left, but $2$ on the right, so they are not equal. –  Arturo Magidin Mar 2 '12 at 5:34
    
Do you know the inequality of the arithmetic and geometric means? –  Gerry Myerson Mar 2 '12 at 5:34
    
@Arturo its not (3/4)^x .Its 3/ (4)^x. –  vikiiii Mar 2 '12 at 5:37
    
@vikiii: That's the problem with not formatting your post correctly. –  Arturo Magidin Mar 2 '12 at 5:41
    
@GerryMyerson its not (3/4)^x .Its 3/ (4)^x. –  vikiiii Mar 2 '12 at 5:42
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2 Answers

up vote 3 down vote accepted

$$4^{x} + 4^{1-x} = (\sqrt{4^x} - \frac{2}{\sqrt{4^x}})^2 + 4 \ge 4$$

with equality occuring when $\sqrt{4^x} = \frac{2}{\sqrt{4^x}}$ and so $x = \frac{1}{2}$.

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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.

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