How do you find the domain and range of $\frac{x}{x^{2} + 1}$ $x/(x^2+1)$ ?

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Answer 1 · 2015-09-18T21:13:21

The domain is $R$ $R$ and the range is $[- \frac{1}{2}, \frac{1}{2}]$ $[-1/2,1/2]$

The domain is $R$ $R$ the set of all reals because $1 + x^{2} > 0$ $1+x^2>0$ .

To find the range we have that

$y = \frac{x}{x^{2} + 1} \Rightarrow y (x^{2} + 1) = x \Rightarrow y x^{2} + x + y = 0 =$ $y=x/(x^2+1)=>y(x^2+1)=x=>yx^2+x+y=0=$

this is trinomial with respect to x hence

$1^{2} - 4 y \cdot y \geq 0 \Rightarrow 1 - 4 y^{2} \geq 0 \Rightarrow - \frac{1}{2} \leq y \leq \frac{1}{2}$ $1^2-4y*y>=0=>1-4y^2>=0=>-1/2<=y<=1/2$

How do you find the domain and range of xx2+1x/(x^2+1)?