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Let $\mathrm{R}$ be the set of real numbers and let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be a
function such that $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{2}}$. What is the range of $\mathrm{f}$ ?
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function such that $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{2}}$. What is the range of $\mathrm{f}$ ?
Solution:
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Verified Answer
The correct answer is:
$[0,1)$
$\because \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{2}}{1+\mathrm{x}^{2}}$
Since, numerator $ < $ denominator $\mathrm{f}(\mathrm{x}) < 1$ for all values of $\mathrm{x}$ (negative or positive) and $\mathrm{f}(\mathrm{x})=0$ for $\mathrm{x}=0$
So, range of fis $[0,1)$.
Since, numerator $ < $ denominator $\mathrm{f}(\mathrm{x}) < 1$ for all values of $\mathrm{x}$ (negative or positive) and $\mathrm{f}(\mathrm{x})=0$ for $\mathrm{x}=0$
So, range of fis $[0,1)$.
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