Epsilon - Delta Problem with polynomial: $\lim\limits_{x\to -3}-4x^2 -40x -96 =-12$

Question

Given f(x) = $-4x^2 -40x -96$ and $\lim\limits_{x\to -3} =-12$, find $\delta$ given $\epsilon$ = 1/6. So far, I have added 12 to f(x) and factorized this into $\lvert-4\rvert$ $\lvert x+3 \rvert$ $\lvert x+7\rvert$. I then converted the absolute value of -4 to just 4 and divided both sides by 4, giving $\lvert x+3 \rvert$ $\lvert x+7\rvert$ $\le$ $\epsilon$ = 1/24. However I'm unsure where to proceed from here? Any ideas?

Answer 1

Cheating a little:

The function is smooth and does not take huge values around $x=-3$. We can risk a sufficiently small value, say $\delta=0.001$.

Then $f(-3-0.001)+12=0.015996$ and $f(-3+0.001)+12=0.016004$, which satisfy $\epsilon$.

But we have to show that $f$ is monotonic in that range, which is true as we are on the same side of the unique maximum, at $x=-5$ (notice $f(x)=-4(x+5)^2+4$).

---

If you need the largest possible $\delta$, you have to solve the two quadratic equations

$$-4x^2-40x-96=-12\pm\frac16$$ for the roots closest to $-3$ and check that the range so obtained does not contain an extremum. The answer is the smallest of $r_++3, r_-+3$.