Question

函数y=2x方+1分之2x方的值域为多少

Answer 1

y=2x/(2x^2+1).
(2x^2+1)y=2x.
2yx^2+y-2x=0.
2yx^2-2x+y=0.(1).
判别式△=(-2)^2-4*2y*y.
=4-8y^2.
∵方程(1)有实数解,∴判别式△≥0.
∴4-8y^2≥0,y^2≤1/2.
|y|≤√2/2.
∴-√2/2≤y√2/2.----即为所求函数y的值域.