Method 1: g (x) = | f' (x) | = |-(x-b) 2+B2+c | [2 refers to the square]. (1)|b| > at 1, it is known from (2) that m >: 2; (2) When | b | is less than or equal to 1, the symmetry axis x=b of the function y=f'(x) lies in the interval [- 1, 1]. At this time, m = max {g (1), g (-65438). [a]: if b is within [- 1, 0], then f' (- 1) is greater than or equal to f'( 1) is less than or equal to f'(b), so g (- 1) is less than or equal to max {g (650). F'(b)} is greater than or equal to1/2 (| f' (1)+| f' (b) |) is greater than or equal to1/2 | f' (1)-f' (b) = 650. Then f'( 1) is less than or equal to f' (- 1) is greater than or equal to f'(b), so g( 1) is less than or equal to max {g (- 1), g (b)}, so m = max {| f}. |f'(b)|} is greater than or equal to1/2 (| f' (-1) | f' (b) is greater than or equal to1/2 | f' (-1)-f' (b