Question

利用导数求函数的极值设函数f(x)=x^2e^(x-1)+ax^3+bx^2已知x=-2 和x=1为f(x)的极值点）（

Answer 1

(1)f'(x) = 2x*e^(x-1) + x^2*e^(x-1) + 3ax^2 + 2bx

极值点的导数值一定为0

所以f'(-2) = -4e^(-3) + 4e^(-3) + 12a - 4b = 12a - 4b = 0

f'(1) = 2 + 1 + 3a + 2b = 0

解得a = -1/3,b = -1

(2)f'(x) = 2x*e^(x-1) + x^2*e^(x-1) - x^2 - 2x = (x^2 + 2x)[e^(x-1) -1]

所以极点有:x = 0,x = -2,x = 1

当x>1时,f'(x)>0

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