已知函数f(x)的定义域为[1,2],求y(x)=(x+1)的定义域
3个回答

1.令t=x+1因为x属于R,得t也属于R则y=f(t)值域为[1.2] 2.应该是求定义域吧由题意,得-2<=根号x<=2,得0<=x<=4  定义域为[0,4]由-2<=x^2<=2,得-根号2<=x<=根号2 定义域为[-根号2,根号2]3.由题意,得-1<=x^2-1<=3,得-2<=x<=2  定义域[-2,2]由-2<=1-3x<=2,得-1/3<=x<=1  定义域为[-1/3,1]4.由题意,得kx^2+4kx+3不等于0得(4k)^2-4*3*k=4k(4k-3)<0得0<k<3/4