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求该函数的间断点,并判断其类型.f(x)=arctan(1/x^2-3x+2)
人气:297 ℃ 时间:2020-03-17 08:24:44
解答
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
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