Question

高二数学19

Answer 1

19、
(1) m=3
f(x)=1/3x^3+x^2-3x+1
f'(x)=x^2+2x-3
=(x+3)(x-1)
f'(x)=0
(x+3)(x-1)=0
x=-3或者x=1
f(1)=1/3×1^3+1^2-3×1+1
=-2/3
f(0)=1
f(3)=1/3×3^3+3^2-3×3+1
=10
最小值：f(x)min=-2/3
最大值：f(x)max=10
(2) f(x)=1/3x^3+x^2-mx+1
f'(x)=x^2+2x-m
f'(x)>0
x^2+2x-m>0
(x+1)^2>m+1
x+1>m+1
x>m
或者x+1<-(m+1)
x<-m-2
∵f(x)在(-2,+∞)上单调递增
∴m<=-2