What values can $\max(\alpha)$ and $\min(\alpha)$ take? In $\{0,0.2,0.4,0.6,0.8,1\}$-valued logic, define $\\f(\alpha)=\min\{\bar{v}(\alpha): v \text{ is some truth valuation on the atoms}\}$. $g(\alpha)=\max\{\bar{v}(\alpha): v \text{ is some truth valuation on the atoms}\}$.
Suppose the connectives evaluate as follows: $\bar{v}(\alpha \wedge\beta) = \min\{\bar{v}(\alpha),\bar{v}(\beta)\}, \bar{v}(\alpha \vee\beta) = \max\{\bar{v}(\alpha),\bar{v}(\beta)\}, \bar{v}(\neg \alpha) = 1-\bar{v}(\alpha)$.
Given a wff $\alpha$, assume it only contains connectives $\neg, \vee,\wedge$, what are the possible values that $f(\alpha)$ and $g(\alpha)$?
I know $f$ can take:
0 ($A \wedge \neg A$, where A=0), $0.6$(e.g. the possible truth values of $A\vee \neg A$ is $1, 0.8, 0.6$, so $f(A\vee \neg A)$ is the min of the three, which is $0.6$).
$g$ can take $1$ ($A\vee \neg A$, A=1), $0.4$ ($A\wedge \neg A$, $A=0.4$ or $0.6$).
Somehow the answer is these are all the values in the ranges of $f$ and $g$. Why?
 A: Here's one way to show this. First, your logic has the following property:
(i) If a formula $\alpha$ takes value $\le 0.4$ somewhere, then it also takes value $0$ somewhere.
Why is this true? For any valuation $v$, consider the valuation $v'$ defined via $v'(x)=\lfloor v(x)+0.5\rfloor\in\{0,1\}$. In other words, $v'$ makes $v$ into a boolean valuation by setting the splitting point at $0.5$. Then by induction on formulas, you can show that $v'(\alpha)=\lfloor v(\alpha)+0.5\rfloor$. Hence if $v(\alpha)\le 0.4$, then $v'(\alpha)=\lfloor v(\alpha)+0.5\rfloor\le\lfloor 0.9\rfloor=0$.
From (i) it follows that $f$ cannot take any value in $\{0.2,0.4\}$.
By a similar argument you can show that
(ii) If a formula $\alpha$ takes value $\ge 0.6$ somewhere, than it also takes value $0.6$ somewhere.
(hint: project all truth values $\ge 0.6$ to $0.6$ and all truth values $\le 0.4$ to $0.4$)
From (ii) it follows that $f$ cannot take a value in $\{0.8,1\}$.
Finally, the possible values for $g$ can be inferred from the above observations, as $g(\alpha)=1-f(\lnot\alpha)$. E.g. $g$ cannot take value $0.2$ as otherwise $f$ would have to take value $0.8$ somewhere, which is not possible.
