Proving the Non-one-to-one Property of f(x)=x^3-3x^2+2x on (-k,k)

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Homework Help Overview

The discussion revolves around proving that the function f(x) = x^3 - 3x^2 + 2x is not one-to-one on the interval (-infinity, +infinity) and finding the largest value of k such that f is one-to-one on the interval (-k, k).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the derivative f'(x) = 3x^2 - 6x + 2 and its implications for the function's monotonicity. Some question the positivity of the derivative and its role in determining one-to-one behavior. Others discuss the discriminant of the derivative to assess where it changes sign and how that relates to the original function's behavior.

Discussion Status

The discussion is ongoing, with participants examining the conditions under which the function may be one-to-one. Some guidance has been offered regarding the use of the quadratic formula to find critical points, which may help identify intervals of monotonicity.

Contextual Notes

There is a focus on the derivative's sign and the implications of its critical points for the function's one-to-one property. Participants are also considering the constraints of the problem, particularly the need to find the largest k for the specified interval.

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Homework Statement


show f(x)=x^3-3x^2+2x is not one to one on (-infinity,+infinity)


Homework Equations



finding the largest value of k such as f is one to one on interval (-k,k)

The Attempt at a Solution


i can get f`(x)=3x^2-6x+2 but it is positive so f(x) should be one-to-one
how to prove it
 
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f(x)=3x^2-6x+2 isn't positive. f(1)=-1.
 
Last edited by a moderator:
I just automatically looked at the discrimant for 3x2- 6x+ 2:
[tex]\sqrt{6^2- 4(3)(2)}= \sqrt{36- 24}= \sqrt{12}[/tex]
Since that is a real number, the 3x2- 6x+2 changes sign, and the original function changes "direction", at two places.
 
HallsofIvy said:
I just automatically looked at the discrimant for 3x2- 6x+ 2:
[tex]\sqrt{6^2- 4(3)(2)}= \sqrt{36- 24}= \sqrt{12}[/tex]
Since that is a real number, the 3x2- 6x+2 changes sign, and the original function changes "direction", at two places.

thanks but how can i do this
largest value of k such as f is one to one on interval (-k,k)
 
A function f(x) is one-to-one as long as its derivative does not change sign- and a continuous derivative, such as the derivative of any polynomial, can change sign only where the derivative is 0.

Solve 3x2- 6x+ 2= 0, say by using the quadratic formula. Those 2 values will give 3 intervals on which the function is one to one. One of them contains the an interval of the form (-k, k).
 

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