Question

y=（x-1）^3（x-2）^4（x-3）^5求拐点数和极值点。。。用图像法或者画数轴来做

Answer 1

y=f(x)=(x-1)³(x-2)⁴(x-3)⁵

f'(x)=3(x-1)²(x-2)⁴(x-3)⁵+4(x-1)³(x-2)³(x-3)⁵+5(x-1)³(x-2)⁴(x-3)⁴

      =(x-1)²(x-2)³(x-3)⁴[3(x-2)(x-3)+4(x-1)(x-3)+5(x-1)(x-2)]

      =(x-1)²(x-2)³(x-3)⁴(12x²-46x+40)

驻点：x₁=1 x₂=2 x₃=3 x₄=5/2 x₅=4/3

f''(x)=2(x-1)(x-2)³(x-3)⁴(12x²-46x+40)+3(x-1)²(x-2)²(x-3)⁴(12x²-46x+40)+

         4(x-1)²(x-2)³(x-3)³(12x²-46x+40)+(x-1)²(x-2)³(x-3)⁴(24x-46)

       =(x-1)(x-2)²(x-3)³[2(x-2)(x-3)(12x²-46x+40)+3(x-1)(x-3)(12x²-46x+40)+

          4(x-1)(x-2)(12x²-46x+40)+(x-1)(x-2)(x-3)(24x-46)]

令g(x)=2(x-2)(x-3)(12x²-46x+40)+3(x-1)(x-3)(12x²-46x+40)+

          4(x-1)(x-2)(12x²-46x+40)+(x-1)(x-2)(x-3)(24x-46)

从图像上看，g(x)有4个零点，且g(1)、g(2)、g(3)均≠0

∴拐点数为7个。

∵f''(1)、f''(2)、f''(3)=0

∴极值点：x=5/2 x=4/3