Let f be a differential function such that fx=f2-x   and   gx=f1+x then

We  have fx=f2-xReplacing x  by  f1+x       f1+x=f1-xHence  graph  of fx  is  symmetrical about   the  line  x=1Also gx=f1+x=f1-x=g-x∴  gx   is  an  even  functionFurther   f'1+x=-f'1-x∴   f'1=-f'1∴  f'1=0