Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a function $$ F(x)= \frac{x^3 - 14x^2 + 7x + 203}{(x-3)(8-x)} $$

I need to use Newton's Method to find the max interval such that a number of constraints are valid.

• $3 < a < b < 8$,

• $f\in C^2[ a, b ]$,

• $f\left( \frac{2}{3}a + \frac{1}{3}b\right)f\left(\frac{1}{3}a+\frac{2}{3}b\right) < 0$,

• $f'(x)\neq 0$ for all $x \in [ a, b ]$,

• $|f(x) f''(x)| < [f'(x)]^2$ for all $x \in (a, b)$.

I have used Newtons Method to discover that the approx root on this function is: $5.22520933956314404$

With some research, i have noted that if $e = \frac{1}{3}(b-a)$, $f$ has a root in $[a+e, b-e]$;

Using this i have verified all of the conditions in my question hold. And they do.

i have used: $a=4.66$ $b=6.33$

Question: How can i know, and prove that $[a+e, b-e]$ is the largest possible interval between $(3, 8)$?

I can provide my script if anyone is interested.

note*: Sorry for formatting, I'm still trying to figure it all out

share|cite|improve this question
What does f C2[ a, b ] mean? Do you mean $f \in [a,b]$? – user2468 Apr 23 '12 at 20:16
Also does x 2 and f'(x)2 mean power 2? i.e. $x^2$ and $f'(x)^2$? – user2468 Apr 23 '12 at 20:19
I do not think this equation was properly translated to LaTeX by @JoeJohnson: |f(x) f''(x)| < f'(x)2 for all x in (a, b). The 2 might be an exponent! – user2468 Apr 23 '12 at 20:21
@J.D.: I will put it in as a power of $2$. The OP can give us guidance, maybe? – Joe Johnson 126 Apr 23 '12 at 20:24
Where can i read about the formatting stuff? Everything above looks correct, thanks for formatting help – KevinCameron1337 Apr 23 '12 at 20:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.