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Find the domain and range of the real function f(x) = x^2 /1+x^2. |
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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)=1and alsof′(x)=2x(1+x2)2f′(x)=2x(1+x2)2so 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). |
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