高数中的拐点.y=(x-1)²(x-3)² 拐点数y=(x-1)²(x-3)² 拐点数y'=4(x-1)(x-2)(x-3)y''=4(3x²-12x+11)解得x1和x2.x1=2+ 根号下3/3 x2=2- 根号下3/3由y''=12(x-x1)(x-x2) 显然y''在x1,x2两侧都变号,所以有两个
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![高数中的拐点.y=(x-1)²(x-3)² 拐点数y=(x-1)²(x-3)² 拐点数y'=4(x-1)(x-2)(x-3)y''=4(3x²-12x+11)解得x1和x2.x1=2+ 根号下3/3 x2=2- 根号下3/3由y''=12(x-x1)(x-x2) 显然y''在x1,x2两侧都变号,所以有两个](/uploads/image/z/6937215-15-5.jpg?t=%E9%AB%98%E6%95%B0%E4%B8%AD%E7%9A%84%E6%8B%90%E7%82%B9.y%3D%28x-1%29%26%23178%3B%28x-3%29%26%23178%3B+%E6%8B%90%E7%82%B9%E6%95%B0y%3D%28x-1%29%26%23178%3B%28x-3%29%26%23178%3B+%E6%8B%90%E7%82%B9%E6%95%B0y%27%3D4%28x-1%29%28x-2%29%28x-3%29y%27%27%3D4%283x%26%23178%3B-12x%2B11%29%E8%A7%A3%E5%BE%97x1%E5%92%8Cx2.x1%3D2%2B+%E6%A0%B9%E5%8F%B7%E4%B8%8B3%2F3+x2%3D2-+%E6%A0%B9%E5%8F%B7%E4%B8%8B3%2F3%E7%94%B1y%27%27%3D12%28x-x1%29%28x-x2%29+%E6%98%BE%E7%84%B6y%27%27%E5%9C%A8x1%2Cx2%E4%B8%A4%E4%BE%A7%E9%83%BD%E5%8F%98%E5%8F%B7%2C%E6%89%80%E4%BB%A5%E6%9C%89%E4%B8%A4%E4%B8%AA)
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高数中的拐点.y=(x-1)²(x-3)² 拐点数y=(x-1)²(x-3)² 拐点数y'=4(x-1)(x-2)(x-3)y''=4(3x²-12x+11)解得x1和x2.x1=2+ 根号下3/3 x2=2- 根号下3/3由y''=12(x-x1)(x-x2) 显然y''在x1,x2两侧都变号,所以有两个
高数中的拐点.y=(x-1)²(x-3)² 拐点数
y=(x-1)²(x-3)² 拐点数
y'=4(x-1)(x-2)(x-3)
y''=4(3x²-12x+11)
解得x1和x2.x1=2+ 根号下3/3 x2=2- 根号下3/3
由y''=12(x-x1)(x-x2) 显然y''在x1,x2两侧都变号,所以有两个拐点
哪里显然的?看不懂
另外求拐点数,如何找拐点?是让y''=0
高数中的拐点.y=(x-1)²(x-3)² 拐点数y=(x-1)²(x-3)² 拐点数y'=4(x-1)(x-2)(x-3)y''=4(3x²-12x+11)解得x1和x2.x1=2+ 根号下3/3 x2=2- 根号下3/3由y''=12(x-x1)(x-x2) 显然y''在x1,x2两侧都变号,所以有两个
方法一:求y''=0的点,y''在该点两侧都变号,所以该点是拐点
方法二:y''=0的点处y'''≠0,该点是拐点