Show continuity of quadratic function using epsilon delta criterion

Question

I want to show using the $\epsilon$-$\delta$ criterion that the function $f(x)=-12x^2+17x+14$ is continuous on $\mathbb{R}$.

I have done the following:

Let $\epsilon>0$ be arbitrary.

For all $x$ with $|x-x_0|<\delta$ we have that \begin{align*}|f(x)-f(x_0)|&=|(-12x^2+17x+14)-(-12x_0^2+17x_0+14)|\\ & =|-12(x^2-x_0^2)+17(x-x_0)|\\ & <12|x^2-x_0^2|+17|x-x_0|\\ & =12|(x-x_0)(x+x_0)|+17|x-x_0|\\ & =12|x-x_0|\cdot |x+x_0|+17|x-x_0|\\ & =(12\cdot |x+x_0|+17)\cdot |x-x_0|\\ & <(12\cdot |x+x_0|+17)\cdot \delta \\ & =(12\cdot |x-x_0+x_0+x_0|+17)\cdot \delta\\ & <(12\cdot |(x-x_0)+2x_0|+17)\cdot \delta\\ & <(12\cdot |x-x_0|+12\cdot |2x_0|+17)\cdot \delta\\ & <(12\cdot \delta+12\cdot |2x_0|+17)\cdot \delta\\ & =12\cdot \delta^2+(24\cdot |x_0|+17)\cdot \delta\end{align*} We are looking for a $\delta$ such that \begin{align*}12\cdot \delta^2+(24\cdot |x_0|+17)\cdot \delta<\epsilon&\Rightarrow 12\cdot \delta^2+(24\cdot |x_0|+17)\cdot \delta-\epsilon<0 \\ & \Rightarrow \delta^2+\frac{24\cdot |x_0|+17}{12}\cdot \delta-\frac{\epsilon}{12}<0\end{align*} We get the negative sign between if $\delta$ is between the two roots of the equation $\delta^2+\frac{24\cdot |x_0|+17}{12}\cdot \delta-\frac{\epsilon}{12}=0$.

The roots are $$\delta_{1,2}=-24|x_0|-17\pm \sqrt{576|x_0|^2+816|x_0|+48\epsilon+289}$$ We have to consider also that $\delta$ must be positive, therefore we are looking for a $\delta$ in the interval $\left (0, -24|x_0|-17+\sqrt{576|x_0|^2+816|x_0|+48\epsilon+289}\right )$ for example $\delta=\frac{-24|x_0|-17+ \sqrt{576|x_0|^2+816|x_0|+48\epsilon+289}}{2}$.

Therefore, we have:

Let $\epsilon>0$ there there is a $\delta$, for example $\delta=\frac{-24|x_0|-17+ \sqrt{576|x_0|^2+816|x_0|+48\epsilon+289}}{2}$ that for all $x\in \mathbb{R}$ with $|x-x_0|<\delta$ it holds that $|f(x)-f(x_0)|<\epsilon$.

Is everything correct? Could I improve something?

Answer 1

You're being much more explicit with everything than you need to. There is really no need to find such a long an explicit expression for $\delta$. What matters is not what $\delta$ is, but that such a $\delta$ exists. So let's look at when you get to

$$\delta^2+\frac{24\lvert x_0\rvert+17}{12}\delta-\frac{\varepsilon}{12}<0$$

for example. At this point, you try to solve for some explicit $\delta$, but there is no need to. See if we rearrange the equation, it becomes

$$\delta\left(12\delta+24\lvert x_0\rvert+17\right)<\varepsilon.$$

Let's force $\delta\leq1$ for convenience (this is a very common trick to make things easier). Then from the above we can write

$$C\delta<\varepsilon$$

for some $C>0$ (if you want, an explicit expression for such $C$ is available from the above). Now from this point you can trivially just choose any $\delta<\frac{\varepsilon}{C}$. So for example, at this point, choosing

$$\delta=\min\left\{\frac{\varepsilon}{C+1},1\right\},$$

you are done. The argument becomes much quicker, much neater, and there is no need for these really long unnecessary expressions.

Now this was just an example of one place in your argument where you can simplify things, and I suspect you can probably shorten the argument even more by doing these kinds of "tricks" a bit earlier on.

Hopefully this provides some good helpful insight for tackling future problems like this!

Answer 2

I think it is much easier to brake it to three parts. Take

$\delta=\dfrac{1}{2}min\left\{\dfrac{\epsilon}{51},\dfrac{\epsilon}{36(2|x_{0}|+1)},\,1 \right\}$. Then $|x-x_{0}|<\,\delta\ \,\, $ implies

$|x-x_{0}|<\dfrac{\epsilon}{36(2|x_{0}|+1)}$<$\dfrac{\epsilon}{36|x+x_{0}|}$ (since $|x-x_{0}|<1$ implies $|x|<1+|x_{0}|$).

Thus, $12|x^{2}-x_{0}^{2}|$<$\dfrac{\epsilon}{3}$. $(1)$

Now since $|x-x_{0}|<\dfrac{\epsilon}{51}\,\,$ we get $17|x-x_{0}|<\dfrac{\epsilon}{3}$ $(2)$ and obviously $|14-14|<\dfrac{\epsilon} {3}$ $(3)$.

Adding together $(1),(2),(3)$ we obtain

$|-12x^{2} +17x+14-(-12x_{0}^{2}+17x_{0}+14|<\,\epsilon$