7
$\begingroup$

$f(x)=4x^2 +x +3$ 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

$\delta= .000111$ 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! $\delta=0.000130$

  • 0
    $\epsilon=0.003$2011-09-03

3 Answers 3

3

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
  • 0
    0.0001304... looks like the correct answer. It is worth noting that $f'(3)=23$ and so you would expect something close to $0.003/23 \approx 0.0001304$2011-09-03
3

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.

  • 0
    @BKaylor Please see my answer below.2011-09-03
3

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 ?" $a,b$ 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.

  • 1
    @BKaylor exactly. Just put the coefficients of the quadratic equation $a=4, b=23, c= -0.003$ into a calculator and you're done2011-09-03