解答题-|||-dfrac (1)(x+1), lt 0-|||-讨论函数 f(x)= https:/img.cdnjtzy.com/zyb_c686e89baf5b90fbb60c97fb998eeb53.jpg-x, leqslant xleqslant 2 的连续性.-|||-dfrac (sin (x-3))(x-3),xgt 2