# 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! $\delta=0.000130$

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I will guess that I go wrong by assuming x is between -4 and -2, because that assumption completely defines my answer and I must be misusing it. –  BKaylor Sep 3 '11 at 18:23
Is epsilon 0.03 or 0.003? –  user38268 Sep 3 '11 at 21:59
$\epsilon=0.003$ –  BKaylor Sep 3 '11 at 22:22

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
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I understand that with a graphing tool I could easily check the intersections of those those values with the function but I can not use a graphing calculator. I think that my trouble is in finding those values algebraically. –  BKaylor Sep 3 '11 at 18:46
@BKaylor: You know you could use the quadratic formula for the first item of Henry's answer, yes? –  Ｊ. Ｍ. Sep 3 '11 at 19:07
I don't see how the quadratic formula would help here. Where would 36.003 and 35.997 be used? –  BKaylor Sep 3 '11 at 19:38
@BKaylor: Write the first expression in the first line of Henry's hint as an explicit equation. Then gather all the terms on the same side. Then you'll find your quadratic for one side, and if you solve that ... then do the second half ... –  Mark Bennet Sep 3 '11 at 20:14
Okay, I got x-values of 2.750130431, -3.000130431, 2.749869563, -2.999869563. Those give me a $\delta$ of 0.000130, and I can see that $f'(x)<0$ all around -3. Is this enough information to conclude that $\delta=0.000130$? –  BKaylor Sep 3 '11 at 21:37

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, $$|x + 3||4x - 11| \leqq |x + 3|(4|x| + 11) < 27|x + 3|.$$ 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.

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By "triangle inequality" I could also mean "ability to look at the graph of $4x - 11$ near $x = -3$". –  Dylan Moreland Sep 3 '11 at 19:20
I don't see how the bound or the inequality works, because if I'm finding a maximum of delta, it seems like there is some specific value of x which gives the maximum of delta for the given epsilon, 0.003. Don't I ultimately have to relate delta to epsilon? –  BKaylor Sep 3 '11 at 19:33
Sorry, do you want to find the best $\delta$ of them all? That is, do you want a $\delta$ such that a $\delta_1 > \delta$ doesn't work in your statement? –  Dylan Moreland Sep 3 '11 at 19:36
And yes, $\delta$ should depend on $\varepsilon$ somehow. That's the part I left for you. –  Dylan Moreland Sep 3 '11 at 19:38
I was under the impression that $0<|x+3|<\delta$ implied that I needed to find the maximum $\delta$. Now that I look at it again, maybe I just need any $\delta$ for which $f(x+\delta)| < |L+\epsilon|$. Do you think that is the case? Then I could see some flexibility in the x value which will ultimately define $\epsilon$. I now see in your inequality that by plugging in the maximum value of x (4) to the factored-out portion of the function gives 27|x+3| which will lead to 0.003/27 for $\delta.$ How did we decide that 0<|x+3|<1? –  BKaylor Sep 3 '11 at 19:52

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.

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Thank you for your answer! I appreciate how in the end you only have to solve one quadratic equation. Is that how you solved for delta in the last step? –  BKaylor Sep 3 '11 at 22:37
@BKaylor exactly. Just put the coefficients of the quadratic equation $a=4, b=23, c= -0.003$ into a calculator and you're done –  user38268 Sep 3 '11 at 23:00