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