If the graph of non constant function is symmetrical about the point $(3,4)$. Then the value of $\sum^{6}_{r=0}f(r) +f(3)$ is
Try: here $f(3)=4$. I did not understand how to find $f(0) f(1),f(2),f(3),f(4),f(5),f(6)$.
Could some help me to solve it, Thanks
Symmetry about a point is not the same as symmetry about the line $x=3$. Symmetry about $(3,4)$ means that $$f(0) + f(6) = 2f(3)$$ $$f(1) + f(5) = 2f(3)$$ and $$f(2) + f(4) = 2f(3)$$
Therefore $\sum_{r=0}^{6}f(r)+f(3) = 8f(3)=32$