Is $f(x) = g(x) ⊕ g(\bar{x})$ a one-way function where $g(x)$ is a OWF? Let $g$ be a one-way function, and let $f(x) = g(x) ⊕ g(\bar{x})$ (where the bar over x denotes bitwise complement). Is $f$ necessarily a one-way function?
Expanding on @Mikero's comment above: First, lets assume that a one-way function $h : \{0,1\}^n \to \{0,1\}^n$ exists.

Then assume towards contradiction that $f(x) = g(x) \oplus g(\bar{x})$ is a one-way function if $g$ is a one-way function. Then in particular $f'(x) = h(x) \oplus h(\bar{x})$ must be a one-way function. And further it follows that 
\begin{align}
f''(x) =\;&amp; f'(x) \oplus f'(\bar{x})\\ =\;&amp; (h(x) \oplus h(\bar{x})) \oplus (h(\bar{x}) \oplus h(x))\\ =\;&amp; (h(x) \oplus h(x)) \oplus (h(\bar{x}) \oplus h(\bar{x}))\\ =\;&amp; 0^n
\end{align}

would be a one-way function, which it is clearly not as it is a constant function.