1.2x+1≥0,得x≥-1/2,定义域[-1/2,+∞)y=x+2√(2x+1)即x^2-2(y+4)x+y^2-4=0设f(x)=x^2-2(y+4)x+y^2-4=[x-(y+4)]^2-8y-20当y+4≤-1/2,即y≤-9/2时,要使得f(x)在[-1/2,+∞)有与x轴有交点则f(-1/2)≤0,即(1/4)+(y+4)+y^2-4≤0,得y=-1/2,与y≤-9/2矛盾当y+4&g 展开
1.2x+1≥0,得x≥-1/2,定义域[-1/2,+∞)y=x+2√(2x+1)即x^2-2(y+4)x+y^2-4=0设f(x)=x^2-2(y+4)x+y^2-4=[x-(y+4)]^2-8y-20当y+4≤-1/2,即y≤-9/2时,要使得f(x)在[-1/2,+∞)有与x轴有交点则f(-1/2)≤0,即(1/4)+(y+4)+y^2-4≤0,得y=-1/2,与y≤-9/2矛盾当y+4>-1/2,即y>-9/2时要使得f(x)在[-1/2,+∞)有与x轴有交点则f(y+4)≤0,即-8y-20≤0,得y≥-5/2所以值域[-5/2,+∞)2.x>0,-x<0,f(-x)=x^2+2xf(x)=-f(-x)=-x^2-2xf(1)=-1-2=-33.f(x+2)是偶函数则f(x+2)=f(-x+2)所以f(2.5)=f(0.5+2)=f(-0.5+2)=f(1.5)f(3.5)=f(1.5+2)=f(-1.5+2)=f(0.5)y=f(x)在(0,2)上是增函数f(0.5)<f(1)<f(1.5)即f(3.5)<f(1)<f(2.5) 收起
