If $$f\left( x \right) = x,\,x \leqslant 1,$$    and $$f\left( x \right) = {x^2} + bx + c,\,x > 1,$$       and $$f'\left( x \right)$$  exists finitely for all $$x\, \in \,R$$   then :

Solution :
$$f\left( x \right)$$  is differentiable at $$x=1$$  also.
$$\eqalign{ & \therefore \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 - h} \right) - f\left( 1 \right)}}{{ - h}} \cr & {\text{Now,}}\,\,\mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^2} + b\left( {1 + h} \right) + c - 1}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2} + \left( {2 + b} \right)h + b + c}}{h}. \cr & \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{1 - h - 1}}{{ - h}} = 1. \cr} $$
The two limits can be equal if $$2+b=1,\,b+c=0$$