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Given relation Q on $R$ (real numbers)

$Q = \{(x,y): y = \frac{2}{{x}^2 + 5}\}$

Find the domain and range of Q and if Q is a function.

Domain is straight forward. The domain for this is ALL the real numbers as the denominator will never go undefined.

I do not know how to find the range and function though, can someone guide on this?

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1 Answer 1

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Hint: $x^2+5\ge 5$ for all $x\in\mathbb{R}$ and what happens if $x$ is large?

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  • $\begingroup$ Still can't understand. I am not sure how to apply the hint you provided to find the range - can you provide abit more hint? $\endgroup$ May 7, 2020 at 9:43
  • $\begingroup$ $x^2+5\ge 5$ implies $\dfrac{2}{x^2+5}\le\dfrac{2}{5}$. To find the lower bound, think about what happens when $x$ is large. $\endgroup$
    – A. Goodier
    May 7, 2020 at 9:45
  • $\begingroup$ I assumed I will just sub a value into x. Say x = 500. If it is $x^2 + 5$, then it will be 250005. And if it is $\frac{2}{{x}^2 + 5}$ it is $\frac{2}{250005}$. It feels like the right hand side of the range is $\frac{2}{5}$ but I can't think why. $\endgroup$ May 7, 2020 at 9:56
  • $\begingroup$ $\frac{2}{250005}$ is very close to zero, which suggests $0$ is the lower bound. It is clear that $y>0$ since it is the quotient of positive numbers. The upper bound of $\frac{2}{5}$ is justified in my previous comment. $\endgroup$
    – A. Goodier
    May 7, 2020 at 10:01
  • $\begingroup$ Ah... I was thinking about lower bound earlier that if y = 0, then it will never be $0 = \frac{2}{{x}^2+5}$ since any extremely large number of x will not get you to this 0 point. $\endgroup$ May 7, 2020 at 10:06

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