The sum of the maximum and minimum values of $3{\mathrm{x}}^4-2{\mathrm{x}}^3-6{\mathrm{x}}^2+6\mathrm{x}+4$ in $\left(0, 2\right)$ is

$$\begin{gathered} f\left(x\right)=3{\mathrm{x}}^4-2{\mathrm{x}}^3-6{\mathrm{x}}^2+6\mathrm{x}+4 \\ {f}^1\left(x\right)=12x^3-6x^2-12x+6 \\ {f}^1\left(x\right)=0 \\ (x-1)(x+1)(2x-1)=0 \\ x=1,\frac{1}{2},-1 \\ f\left(1\right)=3-2-6+6+4=5 \\ f\left(\frac{1}{2}\right)=\frac{3}{16}-\frac{1}{4}-\frac{6}{4}+3+4=\frac{87}{16} \\ Maximum value =\frac{87}{16} \\ minimum value=5 \end{gathered}$$