Question

find the domain and range of the real function f(x) = x^2 /1+x^2.​

Answer 1

Answer: Domain : (0,∞),Range : [0,1).

Step-by-step explanation:

With calculus it’s easier: we see that, with f(x)=x2/(1+x2)f(x)=x2/(1+x2),

limx→−∞f(x)=limx→∞f(x)=1limx→−∞f(x)=limx→∞f(x)=1

and also

f′(x)=2x(1+x2)2f′(x)=2x(1+x2)2

so that ff is strictly decreasing over (−∞,0](−∞,0] and strictly increasing over [0,∞)[0,∞). Since f(0)=0f(0)=0, this and the limits above tell us that the range is [0,1).