\[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f'\left( x \right)=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f'\left( x \right)\leftrightarrow b+1=\frac{3}{2}\to b=\frac{1}{2}\] pertanto la funzione diventa \[f(x)=\left\{ \begin{array}{ll} -1+\sqrt{3x^2+1}\quad x\ge 1 \\ \frac{1}{2}\ln x+x \quad x<1 \end{array} \right. \quad .\]
Massimo Bergamini
