Question

求该函数的间断点,并判断其类型.f(x)=arctan(1/x^2-3x+2)

Answer 1

x^2-3x+2 = (x-1)(x-2) = 0 => x=1, x=2

x->1- , 1/( x^2-3x+2) -> +∞, arctan(1/x^2-3x+2) -> π/2

x->1+ , 1/( x^2-3x+2) -> -∞, arctan(1/x^2-3x+2) -> -π/2

=》 x=1 为第一类跳跃间断点

x->2- , 1/( x^2-3x+2) -> -∞, arctan(1/x^2-3x+2) -> - π/2

x->1+ , 1/( x^2-3x+2) -> +∞, arctan(1/x^2-3x+2) -> π/2

=》 x=2 为第一类跳跃间断点