Finding $\delta$ with given $\varepsilon$ for $f(x)=4x^2 +x +3$ and $x\to3$ $f(x)=4x^2 +x +3$ and the limit as x approaches $-3$ of $f(-3)= 36$, Find $\delta$ such that
$0<|x+3|<\delta \longrightarrow |f(x)-36|<.003$
I have tried:
$|(x+3)(4x-11)|<0.003$
$0<|x+3|<\frac{0.003}{|4x-11|}$
Assume $-4<x<-2$ 
$\delta= .000111$ or $0.000158$  Both came back as incorrect. Where have I gone wrong? 
EDIT: Problem solved via Henry's answer. Thank you all for the help! $\delta=0.000130$
 A: Hints: 


*

*Try to solve $f(x)=36.003$ and then $f(x)=35.997$

*Take the values closer to $-3$ 

*Check that $f'(x)$ does not change sign in that range

A: I feel like you've obtained a correct answer through suspicious reasoning. Let me try to spell out what I think your process is. In the penultimate inequality you have two functions $f$ and $g$ and you're trying to find $x$ such that the $f(x) < g(x)$. To do this, you're evaluating $g$ at a point $a$ and looking at $x$ such that $f(x) < a$. This doesn't seem like enough input to solve the problem. Try drawing an example!
One trick that often works is setting some bound on $|x + 3|$ at the outset. Let's consider only $x$ for which $0 < |x + 3| < 1$. Then by a few applications of the triangle inequality,
\begin{equation}
|x + 3||4x - 11| \leqq |x + 3|(4|x| + 11) < 27|x + 3|.
\end{equation}
The upshot is that for each $0 < \delta < 1$, this will be satisfied by all $x$ such that $0 < |x + 3| < \delta$. Can you see how to finish this off?
Edit. And this appears to give the same answer that WebWork (or whatever you have) doesn't like! Sorry about that. I don't see any flaws here, and I crunched some numbers to make sure that I wasn't going crazy, but corrections are welcome.
It's possible that you are meant to find the "best" $\delta$, but that isn't how the problem is stated. I would try to follow Henry's answer.
A: Let us see how you get the answer without using derivatives. What is the definition of continuity?
A function $f(x)$ is said to be continuous at $x=a$ if for all $\epsilon > 0$, there exists a $\delta >0$ such that 
$|x-a|< \delta$ implies that $|f(x) - f(a)| < \epsilon$.
So let $\epsilon > 0$ be given as in your problem ($\epsilon = 0.003$).
Now if $|x + 3| < \delta$, then we see that 
$-\delta < x+3 < \delta

\implies -4\delta - 23 < 4x - 11 < 4\delta - 23. $
The question is, "Now  that we have a bound on $4x - 11$ of the form $a < 4x - 11 < b$, how can I make it into the form $-c < 4x - 11 <c$?" $a,b$ and $c$ are just some real numbers.
The reason I would like to this is in order for me to put a bound on the absolute value of $4x - 11$.
Notice that $4\delta - 23 < 4\delta + 23$, so that 
$-(4\delta + 23) < 4x - 11 < 4\delta - 23 < 4\delta + 23$.
In other words you can be confident now that
$|4x - 11| < 4\delta + 23$.
So if you consider $|f(x) - 36|$, you see that  $|x +3| < \delta$ will lead to the conclusion that
$|f(x) - 36|= |(x+3)(4x - 11)| < \delta(4\delta + 23)$.
If you solve $\delta(4\delta + 23) = \epsilon = 0.003$ for delta, you will get
$\delta = 0.0001304$ or $\delta = -5.7501304$. Since delta is positive, you will only have one choice of delta, which is 0.0001304 and you're done!
No guessing, playing around with numbers, using derivatives and working blindly.
