Let $$f$$ be a function which is continuous and differentiable for all real $$x.$$ If $$f\left( 2 \right) = - 4$$   and $$f'\left( x \right) \geqslant 6$$   for all $$x\, \in \left[ {2,\,4} \right],$$   then :

Solution :
By mean value theorem, there exists a real number $$c\, \in \left( {2,\,4} \right)$$   such that
$$\eqalign{ & f'\left( c \right) = \frac{{f\left( 4 \right) - f\left( 2 \right)}}{{4 - 2}} \cr & \Rightarrow f'\left( c \right) = \frac{{f\left( 4 \right) + 4}}{2} \cr & {\text{Since, }}f'\left( c \right) \geqslant 6,\,\forall \,x\, \in \left[ {2,\,4} \right] \cr & \therefore \,f'\left( c \right) \geqslant 6 \cr & \Rightarrow \frac{{f\left( 4 \right) + 4}}{2} \geqslant 6 \cr & \Rightarrow f\left( 4 \right) + 4 \geqslant 12 \cr & \Rightarrow f\left( 4 \right) \geqslant 8 \cr} $$