Find $\delta$ for $Q(x)$ in terms of $\epsilon$

Question

This is a homework question.

I got all the way to $$|6x+47||x-8|<\epsilon$$ So I can isolate for $|x-8|$.

What do I do then?

Answer 1

From $|6x+ 47||x- 8|< \epsilon$ you can immediately get $|x- 8|< \frac{\epsilon}{|6x+ 47|}$.

Now, you want to be able to find $\delta$ such that $|x- 8|< \delta$ leads to that. That means you want $\delta$ less that $\frac{\epsilon}{|6x+ 47|}$ so we want a lower bound. That fraction will be smallest when it denominator is largest. So how do we put an upper bound on |6x+ 47|? If we first agree to look only at |x- 8|< 1 or -1< x- 8< 1, then -6< 6x< 6 and 41< 6x+ 47< 53. That is, as long as we know |x- 8|< 1 we can be sure that |5x+ 47|< 53. In that case, we know that $\frac{\epsilon}{|6x+ 47|}< \frac{\epsilon}{53}$.

(NOT 103. Are you sure you have copied that correctly?)

Answer 2

If you factor again $(x-8)$ out of $(6x+47)$ you get $$Q(x)-Q(8)=6(x-8)^2+95(x-8)$$

So $\left|Q(x)-Q(8)\right|\le 6|x-8|^2+95|x-8|<6\delta^2+95\delta=\delta(6\delta+95)$

Now if $\delta<\min\left(1,\dfrac{\varepsilon}{101}\right)$

Then $\left|Q(x)-Q(8)\right|<\dfrac{\varepsilon}{101}(6\times 1+95)=\varepsilon$

I don't know why the proposed answer is $\dfrac{\varepsilon}{103}$ but since it is trivially smaller than $\dfrac{\varepsilon}{101}$, then it would fit also.