I am not sure how to find the correct upper and lower limits for this problem. I did find the breakdown of the inequality to be $\frac{5x^2+|2x|+4}{|5(x^2 + 1)|}$. Thanks for any help you can give!
Prove: $\exists M\in\mathbb{R}$ such that $\forall x\in (1,3)$, $\left|\frac{5x^2 - 2x - 4}{5(x^2 + 1)}\right|\leq M$
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proof-explanation
3 Answers
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Note that for $x$ in that range, $5(x^2+1)>10$. Also, $5x^2-2x-4$ is certainly bounded, e.g., by $5\cdot 3^2+2\cdot 3+4$.
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$|\frac {5x^2 - 2x -4}{5(x^2 + 1)}|\le M \iff$
$-M \le \frac {5x^2 - 2x -4}{5(x^2 + 1)} \le M \iff$
$-M*5(x^2 + 1) \le 5x^2 - 2x - 4 \le M*5(x^2 + 1)\Leftarrow$
$10M \le 5x^2 - 2x - 4 \le 10M$.
Now $5 < 5x^2 < 45$ for $1 < x < 3$.
And $-10 < -2x - 4 < -6$ for $1 < x < 3$.
So $-5 < 5x^2 - 2x - 4 < 39$.
So if $M = 3.9$ then
$-10M < -5 < 5x^2 - 2x - 4 < 39 = 10M \implies$
$- M*5(x^2 + 1) \le -10M < 5x^2 - 2x - 4 < 10M \le M*5(x^2 + 1)\implies$
$|\frac {5x^2 - 2x -4}{5(x^2 + 1)}|\le M$.
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Consider the function $f:(1,3) \rightarrow \mathbb{R}$ given by $f(x)=\frac{5x^2 - 2x - 4}{5(x^2 + 1)}$, $\forall x \in (1,3)$. Then f is uniformly continuous on its domain.So, f can be continuously extended to $F:[1,3] \rightarrow \mathbb{R}$. Now F being continuous on a compact domain attains extremums. Hence there is some $M$ such that $| \frac{5x^2 - 2x - 4}{5(x^2 + 1)} | \leq M$.