Help regarding computation of delta value

[DigitalSplendid]

Problem:​

Solution:


Would help if anyone could confirm if the way I have computed delta value correct or not?

Graphical solution using Wolfram Mathematica (optional):

 

[logistic_guy]

Problem:​

View attachment 39101

Solution:
View attachment 39102

Would help if anyone could confirm if the way I have computed delta value correct or not?

Graphical solution using Wolfram Mathematica (optional):
View attachment 39103

You got $\displaystyle \delta = 0.125$ which is wrong.

I think that this is the correct approach by using the hint.

Find the values of $\displaystyle x$ when $\displaystyle x^3 - 4x + 5 = 1.95$
I used WolframAlpha and I got:
$\displaystyle x \approx -2.30696$
$\displaystyle x \approx 1.06163$
$\displaystyle x \approx 1.24534$

Find the values of $\displaystyle x$ when $\displaystyle x^3 - 4x + 5 = 2.05$
$\displaystyle x \approx -2.29857$
$\displaystyle x \approx 0.95578$
$\displaystyle x \approx 1.34279$

Now solve for $\displaystyle \delta$. The possible values for $\displaystyle \delta$ are:
$\displaystyle \delta_1 = |-2.30696 - 1| = 3.30696$
$\displaystyle \delta_2 = |1.06163 - 1| = 0.06163$
$\displaystyle \delta_3 = |1.24534 - 1| = 0.24534$
$\displaystyle \delta_4 = |-2.29857 - 1| = 3.29857$
$\displaystyle \delta_5 = |0.95578 - 1| = 0.04422$
$\displaystyle \delta_6 = |1.34279 - 1| = 0.34279$

The only $\displaystyle \delta$ of those that satisfies the main condition is $\displaystyle \delta_5 = 0.04422$.

Why?

I used WolframAlpha to solve $\displaystyle |f(x) - 2| < 0.05 \rightarrow |x^3 - 4x + 3| < 0.05$.

I got:

$\displaystyle -2.30696 < x < -2.29857$
$\displaystyle 0.95578 < x < 1.06163$
$\displaystyle 1.24534 < x < 1.34279$

If we choose $\displaystyle \delta_1 = 3.30696$, then $\displaystyle 0 < |x - 1| < 3.30696$.
This gives the interval of $\displaystyle x$: $\displaystyle (-2.30696,1) \cup (1,4.30696)$.
This does not satisfy the main condition because $\displaystyle x$ cannot be $\displaystyle 4$.

If we choose $\displaystyle \delta_2 = 0.06163$, then $\displaystyle 0 < |x - 1| < 0.06163$.
This gives the interval of $\displaystyle x$: $\displaystyle (0.93837,1) \cup (1,1.06163)$.
This does not satisfy the main condition because $\displaystyle x$ cannot be $\displaystyle 0.94$.

You can check the rest by yourself to see that they don't satisfy the main condition.

But If we choose $\displaystyle \delta_5 = 0.04422$, then $\displaystyle 0 < |x - 1| < 0.04422$.
This gives the interval of $\displaystyle x$: $\displaystyle (0.95578,1) \cup (1,1.04422)$.
This indeed the only $\displaystyle \delta$ that satisfies the main condition.

 

[blamocur]

You are not likely to get much help when both your hand-writing and the screenshot are unreadable.

What is your answer for $\delta$ ?

 

[jonah2.0]

Beer induced hack reaction follows.

Problem:​

View attachment 39101

Solution:
View attachment 39102

Would help if anyone could confirm if the way I have computed delta value correct or not?

Graphical solution using Wolfram Mathematica (optional):
View attachment 39103

$\displaystyle |(x^3−4x+5)−2| < 0.05$ is equivalent to $\displaystyle −0.05 < (x^3−4x+5)−2 < 0.05$,
which means $\displaystyle 1.95 < x^3−4x+5 < 2.05$. Now $\displaystyle x^3−4x+5 = 1.95$ at x = 1.0616, and $\displaystyle x^3−4x+5 = 2.05$ at x = 0.9558. So $\displaystyle \delta$= min $\displaystyle (1.0616 − 1, 1 − 0.9558) = 0.0442$.