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Can anyone double check my following epsilon delta proof.

I want to prove that the following function is continuous with an Epsilon Delta Argument. $$ f: x \in \mathbb R \mapsto (4x^2+3x+17) \in \mathbb R $$

So I started with $$\left\lvert f(x)-f(y) \right\rvert =$$ $$= \left\lvert (4x^2+3x+17) - (4y^2+3y+17) \right\rvert$$ $$= \left\lvert (4x^2+3x)-(4y^2+3y) \right\rvert$$ $$= 3\left\lvert (4x^2+x)-(4y^2+y) \right\rvert$$ $$= 3\left\lvert (\frac {4}{3}x^2+x)-(\frac {4}{3}4y^2+y) \right\rvert $$ $$= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert$$ $$= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert$$ (Triangel inequality)

$$\le 3 \left(\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 \right\rvert + \left\lvert x-y \right\rvert \right)$$

Note that $\delta \le 1$

$$\le 3 \left(\frac {4}{3}\left\lvert x^2 - y^2 \right\rvert + \delta \right)$$

$$= 3 \left(\frac {4}{3}\left\lvert (x - y) \right\rvert\left\lvert (x +y) \right\rvert + \delta \right)$$

$\delta \le 1$ $$\le 3 \left(\frac {4}{3}\left\lvert x + y \right\rvert \delta + \delta \right)$$ Adding zero in form of y - y $$= 3 \left(\frac {4}{3}\left\lvert x + y - y + y \right\rvert \delta + \delta \right)$$ $$= 3 \left(\frac {4}{3}\left\lvert x - y + 2y \right\rvert \delta + \delta \right)$$ Triangel inequality $$\le 3 \left(\frac {4}{3}(\left\lvert x - y \right\rvert +\left\lvert 2y \right\rvert) \delta + \delta \right)$$ Note that $\delta \le 1$

$$\le 3 \left(\frac {4}{3}(\delta +\left\lvert 2y \right\rvert) \delta + \delta \right)$$ $$= 4\delta^2+4\left\lvert 2y \right\rvert \delta + 3\delta $$ $$= \delta(4\delta+4\left\lvert 2y \right\rvert + 3) $$ Note that $\delta \le 1$ $$\le \delta(4 +4\left\lvert 2y \right\rvert + 3) $$ $$\le \delta(\left\lvert 8y \right\rvert + 7) = \epsilon$$ Therefore $$\delta = \frac {\epsilon}{(\left\lvert 8y \right\rvert + 7)} \;\; with \; \delta \le1$$

Thanks in advance for the help, I really appreciate it. :)

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    $\begingroup$ Your final line is that $\delta$ is expressed in terms of the variable $y$. That is an absolute and unresolvable no-no. $\endgroup$
    – fleablood
    Jan 14 at 17:57
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    $\begingroup$ @fleablood his $y$ is (something like) $x+\delta$, so maybe with some modifications between the last two lines it's fixable? $\endgroup$ Jan 14 at 18:02
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    $\begingroup$ "Why is that a problem?" Because to choice $y$ so that $|x-y| < \delta$ you have to know what $\delta$ is first. If $\delta = $something to do with $y$ you have to know $y$ first which is circular; you have to know what $y$ is before you can choose it. @BenjaminWang "his y is (something like) x+δ, so maybe with some modifications between the last two lines it's fixable? " Maybe. Probably. Conventionally we don't do $\delta$ dependent on $x$ but.... off hand that will prove it is continuous at that $x$ but as $x$ is arbitrary that could be acceptable... I think. $\endgroup$
    – fleablood
    Jan 14 at 18:20
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    $\begingroup$ $|2y| < 2|x| + 2\delta$ so $\delta(8|x| + 4\delta + 7)\le (8|x|+ 4)\delta +7$ ought to do it. I thing. But I'm a bit fussy about saying "Therefore $\delta = ....$" ... that's not really the logic of the proof. Better to say "therefore of we choose a $\delta$ so that $0 < \delta \le \min(..., 1)$ that will be a valid choice and our result will follow" or something like that. $\endgroup$
    – fleablood
    Jan 14 at 18:31
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    $\begingroup$ " Isn't it okay then?" Yes, and that delta will work. And any delta smaller would work. And that is not the largest possible choice of delta. But it is an acceptable delta. To my mind saying "therefore $\delta =$" sounds like you are saying delta must be that value. It'd be more accurate to say "If we choose $\delta$ to be this value that will be sufficient". ... but it's a minor point. $\endgroup$
    – fleablood
    Jan 14 at 18:45
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Your solution looks good. The following steps could be simplified a bit to make it more readable.

\begin{align} \left\lvert f(x)-f(y) \right\rvert &= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert \quad (\textrm{why bother factoring out $3$?})\\ & \le 3 \left(\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 \right\rvert + \left\lvert x-y \right\rvert \right)\\ & \le 3 \left(\frac {4}{3}\left\lvert x^2 - y^2 \right\rvert + \delta \right) \quad (\color{red}{\textrm{for $|x-y|<\delta$} })\\ &= 3 \left(\frac {4}{3}\left\lvert (x - y) \right\rvert\left\lvert (x +y) \right\rvert + \delta \right)\\ &= 3 \left(\frac {4}{3}\left\lvert x - y + 2y \right\rvert \delta + \delta \right)\\ &\le 3 \left(\frac {4}{3}(\left\lvert x - y \right\rvert +\left\lvert 2y \right\rvert) \delta + \delta \right) \quad (\textrm{triangle inequality})\\ &\le 3 \left(\frac {4}{3}(\delta +\left\lvert 2y \right\rvert) \delta + \delta \right)\\ &=4(\delta +\left\lvert 2y \right\rvert) \delta + 3\delta \\ &\le 4(1+\left\lvert 2y \right\rvert) \delta + 3\delta \quad (\delta\le 1)\\ &\le (7+|8y|)\delta \end{align}

Set $$\delta =\min(\frac {\epsilon}{\left\lvert 8y \right\rvert + 7} ,1).$$

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To prove that $f(x)=4 x^2 + 3 x + 17$ is continuous $\forall a\in\mathbb{R}$

$$\underset{x\to a}{\text{lim}}\left(4 x^2+3 x+17\right)=f(a)$$ Equivalently $$\underset{h\to 0}{\text{lim}}\left(4 (a+h)^2+3 (a+h)+17\right)=4 a^2 + 3 a + 17$$ $\forall \varepsilon>0$ we must find a $\delta>0$ such that if $|h|<\delta$ then $$|4 (a+h)^2+3 (a+h)+17-(4 a^2 + 3 a + 17)|<\varepsilon$$ that is $$|8 a h+4 h^2+3 h|<\varepsilon$$

$$0<\delta<\left|\frac{1}{8} \sqrt{64 a^2+48 a+16 e+9}+\frac38-a\right|$$ if $|h|<\delta$ then $|f(a+h)-f(a)|<\varepsilon$

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  • $\begingroup$ Thanks for your post. But could you please verify if or if not my answer is correct. I know that there are different ways like yours but it doe not really help me. $\endgroup$
    – user857338
    Jan 14 at 20:59
  • $\begingroup$ Express $\delta$ as $ \frac {\varepsilon}{7}$. So you prove that $f(x)$ is uniformly continuos on any $[a,b]$ $\endgroup$
    – Raffaele
    Jan 14 at 21:17
  • $\begingroup$ Why can I just leave out the $ (\left \vert 8y \right \vert )^{-1}$? $\endgroup$
    – user857338
    Jan 14 at 21:21

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