This is the second part of a question I asked earlier.
I tried to figure it out but was having trouble even with 4 of my peers.
PART 1 (SOLVED)
A function f is said to symmetric about a point (p,q) if whenever the point (p-x, q-y) is on the graph of f, then the point (p + x, q - y) is also on the graph. Said differently, f is symmetric about a point (p,q) if the line through the points (p,q) and (p+x, q+y) on the graph of y=f(x) intersects the graph at the point (p-x, q-y). Show that a function symmetric about the point (p,q) satifies
f(p-x) + f(p+x) = 2f(p) for all x in the interval of interest.
f(p+x) + f(p-x) = 2f(p)
f(p+x) = q+y
f(p-x) = q-y
f(p+x) + f(p-x) = (q+y) + (q-y)
f(p+x) + f(p-x) = 2q
we know f(p) = q so substituting this in the 2q we get
f(p+x) + f(p-x) =2f(p)
Part 2 UNSOLVED:
Suppose that f is a continuous function on the interval |a,b| and is symmetric about the point ((a+b)/2,f(a+b)/2) (the point whose x coordinate is the midpoint of [a,b] and whose y coordinate is the corresponding point on the graph of f). Using Step 1, show that f satisfies f(x) + f(a+b-x) = 2f(a+b)/2) for all x in [a,b].
The Attempt at a Solution
I made a few attempts but none of them seem to incorporate the first step. Can anyone please help me get started atleast?