更新时间2019-07-15 05:49:35
求解答
y = 4(x+1)/x^2 - 2;
断点,x = 0;x → 0,y → ∞;
lim x → ±∞,4( 1 + 1/x )/x - 2 → -2;
y' = 4 [ x^2 - 2x( x + 1 ) ]/x^4 = 0,x = -2;
y'' = [ 1/x^2 - 2( x + 1 )/x^3 ]' = -2/x^3 - [ 2x^3 - 2( x + 1 ) * 3x^2 ]/x^6
= 2{ [ 2x + 3 ]/x^4 - 1/x^3 } = 0,( 2x + 3 )/x = 1,x = -3;∴ x = -3 是拐点;
y''(-2) = 2{ [ -4 + 3 ]/(-2)^4 - 1/(-2)^3 } = 2{ -1/16 + 1/8 } > 0,
∴ x = -2 是极小点,曲线上凹;
y''(1) = 2{ [ 2 + 3 ]/1^4 - 1/1^3 } = 2{ 5 - 1} > 0,∴ x > 0,曲线上凹;
y(-2) = 4( -2 + 1 )/(-2)^2 - 2 = -3 < -2;
故,x = -2 是极小点,极小值为 -3;
区间 ( -∞,-2 ) ∪ ( 0,∞ ),函数单调减;区间 ( -2,0 ),函数单调增;
有一个拐点,x = -3;
区间 ( -∞,-3 ) 函数上凸;区间 ( -3,0 ) ∪ ( 0,∞ ),函数上凹;
水平渐近线是 y = -2;垂直渐近线是 x = 0 。