Assuming the function is
[tex]f(x)=\begin{cases}2x&\text{for }x\le1\\bx^2+cx&\text{for }x>1\end{cases}[/tex]
For [tex]f(x)[/tex] to be differentiable, it necessarily has to be continuous. For this condition to be met, we need
[tex]\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)[/tex]
[tex]\iff\displaystyle\lim_{x\to1}2x=\lim_{x\to1}(bx^2+cx)[/tex]
[tex]\iff2=b+c[/tex]
For the derivative to exist, the one-sided limits of the derivative must also exist and be equal. We have
[tex]f'(x)=\begin{cases}2&\text{for }x<1\\2bx+c&\text{for }x>1\end{cases}[/tex]
[tex]\displaystyle\lim_{x\to1^-}2=\lim_{x\to1^+}(2bx+c)[/tex]
[tex]\iff2=2b+c[/tex]
Now we solve for [tex]b[/tex] and [tex]c[/tex]:
[tex]\begin{cases}b+c=2\\2b+c=2\end{cases}\implies b=0,c=2[/tex]
