f(x)=2(k+1)x²+4x+3k-2
f(x)有最小值则k+1>0 k>-1
f'(x)=4(k+1)x+4=0
x=-1/(k+1)
f(-1/(k+1))=2/(k+1)-4/(k+1)+3(k+1)-5<0
-2/(k+1)+3(k+1)-5<0
3(k+1)²-5(k+1)-2<0
(3(k+1)+1)((k+1)-2)<0
-1/3<(k+1)<2
0<(k+1)<2
-1
当f(x)＜1时递增求k的范围 不成立
应为
当x＜1时递增求k的范围
f'(x)=4(k+1)x+4=0
x=-1/(k+1)
当k+1>0 则x≥-1/(k+1)f(x)增 无解
当k+1<0 则x≤-1/(k+1)f(x)增
-1/(k+1)>1
k+1>-1
k>-2
取 -2
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