${\displaystyle a:~a\in (-\infty ,\infty )}$
${\displaystyle b:~b>a}$
${\displaystyle c:~a\leqslant c\leqslant b}$

${\displaystyle a\leqslant x\leqslant b}$

${\displaystyle \left\{{\begin{matrix}{\frac {2(x-a)}{(b-a)(c-a)}}&\mathrm {dla\ } a\leqslant x\leqslant c\\&\\{\frac {2(b-x)}{(b-a)(b-c)}}&\mathrm {dla\ } c\leqslant x\leqslant b\end{matrix}}\right.}$

${\displaystyle \left\{{\begin{matrix}{\frac {(x-a)^{2}}{(b-a)(c-a)}}&\mathrm {dla\ } a\leqslant x\leqslant c\\&\\1-{\frac {(b-x)^{2}}{(b-a)(b-c)}}&\mathrm {dla\ } c\leqslant x\leqslant b\end{matrix}}\right.}$

${\displaystyle {\frac {a+b+c}{3}}}$

${\displaystyle \left\{{\begin{matrix}a+{\frac {\sqrt {(b-a)(c-a)}}{\sqrt {2}}}&\mathrm {dla\ } c\!\geqslant \!{\frac {b\!-\!a}{2}}\\&\\b-{\frac {\sqrt {(b-a)(b-c)}}{\sqrt {2}}}&\mathrm {dla\ } c\!\leqslant \!{\frac {b\!-\!a}{2}}\end{matrix}}\right.}$

${\displaystyle c}$

${\displaystyle {\tfrac {a^{2}+b^{2}+c^{2}-ab-ac-bc}{18}}}$

${\displaystyle {\tfrac {{\sqrt {2}}(a\!+\!b\!-\!2c)(2a\!-\!b\!-\!c)(a\!-\!2b\!+\!c)}{5(a^{2}\!+\!b^{2}\!+\!c^{2}\!-\!ab\!-\!ac\!-\!bc)^{\frac {3}{2}}}}}$

${\displaystyle -{\frac {3}{5}}}$

${\displaystyle {\frac {1}{2}}+\ln \left({\frac {b-a}{2}}\right)}$

${\displaystyle 2{\tfrac {(b\!-\!c)e^{at}\!-\!(b\!-\!a)e^{ct}\!+\!(c\!-\!a)e^{bt}}{(b-a)(c-a)(b-c)t^{2}}}}$

${\displaystyle -2{\tfrac {(b\!-\!c)e^{iat}\!-\!(b\!-\!a)e^{ict}\!+\!(c\!-\!a)e^{ibt}}{(b-a)(c-a)(b-c)t^{2}}}}$