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 [imath]\delta[/imath] ?

 

[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 \(\displaystyle (1.0616 − 1, 1 − 0.9558) = 0.0442\).