If f''(0)>0
There is u>0, such that f'(x) is strictly increasing on (-u u). Then (f(u)-f(0))/(u-0) = f'(x1)>f'(0)>f'(x2) = (f(0)-f(-u))/(0-(-u))
where x1 is some number in (0 u), x2 is some number in (-u 0)
defing g(x) = (f(u-x)-f(-x))/u, then g is continuous, g(0)>f'(0)>g(u), by intermdiate value theorem, there exists y in (0,u), such that g(y) = f'(0). Contradiction.
Similarly, f''(0)<0 leads to contradiction.
There is u>0, such that f'(x) is strictly increasing on (-u u). Then (f(u)-f(0))/(u-0) = f'(x1)>f'(0)>f'(x2) = (f(0)-f(-u))/(0-(-u))
where x1 is some number in (0 u), x2 is some number in (-u 0)
defing g(x) = (f(u-x)-f(-x))/u, then g is continuous, g(0)>f'(0)>g(u), by intermdiate value theorem, there exists y in (0,u), such that g(y) = f'(0). Contradiction.
Similarly, f''(0)<0 leads to contradiction.
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