Let f (x) be an odd function on (- ∞, + ∞), f (x + 2) = - f (x). When 0 ≤ x ≤ 1, f (x) = x, then f (7.5) is equal to Why ∵ f (x + 2) = - f (x), ∵ f (x + 4) = f (x) ∵ f (7.5-8) = f (- 0.5)? Answer!

Let f (x) be an odd function on (- ∞, + ∞), f (x + 2) = - f (x). When 0 ≤ x ≤ 1, f (x) = x, then f (7.5) is equal to Why ∵ f (x + 2) = - f (x), ∵ f (x + 4) = f (x) ∵ f (7.5-8) = f (- 0.5)? Answer!

This is a significant problem of periodic function. I will simply push it out, because f (x + 2) = - f (x) replaces x with - x to get f (- x + 2) = - f (- x). Because f (x) = - f (- x) gets f (x + 2) + F (2-x) = 0 gets f (x + 2) = - f (2-x) = f (X-2), replaces X-2 with X to get f (x + 4) = f (x), and then replaces x with x + 4 to get f (x + 8) = f (x)