已知函数f(x)=(2k-3)x^2+2kx+k+2(1)如果当x1时,f(x)单调递减,求实数k的取值范围(2)如果函

(1)f '(x)=2(2k-3)x+2k当x＜0,函数递增则f '(x)=2(2k-3)x+2k＞0即-k/(k-3)＜x＜0得k＞3或k＜0当x＞1,函数递减则f '(x)=2(2k-3)x+2k＜0即-k/(k-3)＞x＞1得3/2＜k＜3综上,k的取值范围：(负无穷大,0)U(3/2,3)U(3,正无穷大)(2)f(x)=2(k-2)[x+k/(2k-6)]?-k?/(2k-6)+k+2过一二三象限,不过第四象限,即与Y轴交于正半轴-k?/(2k-6)+k+2＞0解之,得k＞1+根号11或k＜3