Epsilon-delta Challenge

This is a classical approach to limits. :)

From the problem, we shall extract the important details that could help us answer the question stated. Thus, $0<|x-2|<\delta$

$f(x)=3x^2 +2x+1$ , and

$|f(x)-17|<0.5$

First, we have to evaluate the latter given. $|f(x)-17|<0.5)$

$|3x^2 +2x +1 -17|<0.5$

$|3x^2+2x-16|<0.5$

$|(3x+8)(x-2)|<0.5$

$|3x+8||x-2|<0.5$

Obviously, from the strict inequality from the given, we can say that,

$|x-2|<\delta$

and to get $|3x+8|$

$-\delta<x-2<\delta$

$-3\delta<3x-6<3\delta$

$-3\delta+14<3x+8<3\delta+14$

$|3x+8|<3\delta+14$

Hence,

$|3x+8||x-2|<0.5$

$(3\delta+14)(\delta)<0.5$

$3\delta^2+14\delta-0.5<0$

Using Quadratic formula, we will get

$-4.70211<x<0.03544$

Ofcourse, $0<x<\delta$

Thus,

$0<x<0.03544$

and the nearest value to delta that falls under the continued inequality is $0.03$

In order to ensure $f(x)$ is no more than 0.5 units away from 17 for all $x$ on the interval $(2 - \delta, 2 + \delta),$ we will solve $\left|3x^2 + 2x + 1 - 17\right| < 0.5.$

Solving this inequality yields two intervals: $1.96401 < x < 2.03545$ (the other interval does not capture $x = 2$ , so we ignore it). The disparity from $x = 2$ in this interval is roughly 0.035, so any choice of $\delta$ less than this will be sufficient to respond to Alice's challenge. Out of the given answers, the largest is $\boxed{0.03}.$

An alternative starting from the "obviously" part:

After obtaining the inequality

∣3x+8∣∣x−2∣< 0.5

We can divide both sides of the inequality by∣3x+8∣to get

∣x−2∣< 0.5 /∣3x+8∣

Since we know that

∣x−2∣< δ

We can LET δ = 0.5 /∣3x+8

Knowing that x approaches 2, plug 2 into the new inequality, and we can get:

δ = 0.5 /∣3*2+8∣

=> δ = 0.5 / 14

=> δ = 0.0357

Delta should be less than epsilon. Here three options satisfy and as the question demands the largest among the three, we have 0.03.

We have $|f(x)-17|<0.5$

$|3x^2+2x-16|<0.5$

$|(x-2)(3x+8|<0.5$

As $x$ approaches $2$ , then $1<x<3$

$3<3x<9$

$11<3x+8<17$

Then $|3x+8|<17$ ........(i)

And $|x-2|<\delta$ ........(ii)

We get $|x-2||3x+6|<17 \delta =\epsilon =0.5$

So $\delta =\frac{0.5}{17}=0.029411764705882$

By rounding this number we get $\delta=0.03$ is the answer.

I am lazy so to solve by mental computation, I started with L = 17.5 I used the root formula that I got sqrt(202) at the sqrt (b^2 - 4ac) part to get x. I found that I should approximate so I tried 14^2 =196 to compare sqrt(202). So you can make sqrt(202) equals f(196+6) to use linear approximation for the sqrt part. So you approximate sqrt(202) to 1/(2sqrt(196))*6+14) which is 3/14+14 Back to the root formula, (12+3/14)/6. Is 12/6 + 1/28. Since I wanted to get delta and I have 2 as the first term of x from L, 1/28 is approximation of the answer. 1/28 is 0.03something something...

So 0.03 was reasonable

|3x^2 + 2x + 1 - 17| < .5 |3x^2 +2x -16| <.5 |(3x + 8)| |(x - 2)| < .5 divide both sides by |3x + 8| substitute x with c (2)

f(x)=3x^2 +2x+1

$\frac{df}{dx}$ = 6x + 2

df = (6x + 2)*dx

0.5 = (6 * 2 + 2) * dx (become more accurate if dx approaches 0, and we know that dx is very small)

dx = 0.5/14

dx = 0.035714...

so we got 0.3

sorry I'm still learning english