Since, Rolle's theorem is satisfied in $[-1, 1]$ and corresponding to $x=\frac{1}{2}$ we have  $f(-1)=f(1)$   &   $f^{\text{'}}\left(\frac{1}{2}\right)=0$

So, $-2+b-c=2+b+c$

$\Rightarrow c=-2$

Further $f^{\text{'}}\left(x\right)=6x^2+2bx+c$

Put $x=\frac{1}{2}$

So $f^{\text{'}}\left(\frac{1}{2}\right)=\frac{3}{2}+b+c=0$

$\Rightarrow b=\frac{1}{2}$

Hence, $2b+c=-1$