Proof Check: $4x^2+3x+17$ is continuous by using a $\epsilon$, $\delta$ Argument

Question

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. :)

Answer 1

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

Answer 2

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 $|f(a+h)-f(a)|<\varepsilon$