Simple Quiz

[kean]

Is there a number that is exactly 1 more than its cube?

Show your proof if your answer is yes or no.:dance:

 

[Vadim]

There is just one real number:

\(f(x) = x^3-x+1=0\)
\(f'(x)=3x^2-1=0\)
\(x_1=\frac{1}{sqrt(3)} x_2=-\frac{1}{sqrt(3)}\)
\(f(\frac{1}{sqrt(3)}) = 0.615100\) - minimum
\(f(-\frac{1}{sqrt(3)}) = 1.384900\) - maximum
\(f(-infinity) = -infinity\) it means you have a solution for f(x)=0

such solution is close to
x = -1.3247179574971
with approximation error = .0000000010761969093664

-V-

 

[kean]

Not quite right

Sorry mate. Not quite right. Hint: an undergraduate type question based on Intermediate Value Theorem.

There is just one real number:

\(f(x) = x^3-x+1=0\)
\(f'(x)=3x^2-1=0\)
\(x_1=\frac{1}{sqrt(3)} x_2=-\frac{1}{sqrt(3)}\)
\(f(\frac{1}{sqrt(3)}) = 0.615100\) - minimum
\(f(-\frac{1}{sqrt(3)}) = 1.384900\) - maximum
\(f(-infinity) = -infinity\) it means you have a solution for f(x)=0

such solution is close to
x = -1.3247179574971
with approximation error = .0000000010761969093664

-V-

 

[Andy Nguyen]

\(f(x)=x^3-x+1\)
\(f(0)=1,f(-2)=-5 \Rightarrow f(0)f(-2) < 0\)
By IVT, this shows that f(x) has at least one root in the interval (-2,0)

To find that root, you want to show me why Vadim's number is not right, Calvin ?

 

[kean]

FYI

Andy,

The question did not ask to find the root. I put this question here because I did exactly the same like your answer plus Vadim's answer.
My lecturer raised a good question: Do you need to answer what I do not ask for?

The main issue is understanding of the question. So, do we really understand the market when we do maths or modeling??

I leave it to you. What the question is actually ask for is more important in this aspect. I did not ask to find the root.

Cheers,

 

[kean]

No intention to challenge anyone. This is merely a knowledge sharing post with due respect.
Thanks.