Understanding Breaking Down h'(x) and |x|+|x+1|

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Discussion Overview

The discussion revolves around the analysis of the derivative of a quadratic function, specifically h(x) = 3x^2 - 3, and the breakdown of absolute value expressions, particularly |x| + |x+1|. Participants are exploring how to determine intervals of increase and decrease for the function based on its derivative, as well as the implications of absolute value definitions in different domains.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about analyzing the derivative h'(x) = 3(x+1)(x-1) and how it relates to determining intervals of increase and decrease.
  • Another participant states that for h(x) = 3x^2 - 3, the correct derivative is h'(x) = 6x, identifying a critical point at x = 0 and noting that the function is decreasing for x < 0 and increasing for x > 0.
  • There is a question about how the condition x < -1 leads to the expressions |x| = -x and |x + 1| = -x - 1, with a participant asking for clarification on the meaning of |x|.
  • Some participants suggest that there may be a typo in the derivative provided by the original poster, emphasizing the need to correct it.
  • One participant proposes analyzing when the factors (x-1) and (x+1) are positive to determine when their product is positive, implying a method for finding intervals of increase and decrease.
  • Another participant suggests plotting the parabola to visually identify the vertex and intercepts, offering an alternative approach to understanding the function's behavior.

Areas of Agreement / Disagreement

There is no consensus on the correct derivative of h(x), as some participants assert it should be h'(x) = 6x while others reference a different factorization. The discussion remains unresolved regarding the correct interpretation of the absolute value expressions and their implications.

Contextual Notes

Participants are working with different interpretations of the derivative and the absolute value functions, leading to confusion about the conditions under which certain expressions hold true. The discussion highlights the importance of clarity in mathematical definitions and the potential for typographical errors to influence understanding.

semidevil
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i'm trying to do analysis, but the stupid case by case part is confusing me and it's keeping me away from doing calculus.

ex:
h(x) = 3x^2 - 3
h'(x) = 3(x+1)(x-1)

how did they break down h'x to analyze it?

like, for x <= -1, --> h'x >= 0 --> h is increasing
-1<x<1 --> h'x <= 0 --> h is decreasing..

etc etc...


stuff like that...how did they get h'x is greater or lesser then 0 and increasing decreasing?


also

|x| + |x+1|

how did they get x < -1 implies |x| = -x and |x + 1| = -x - 1?


i'm sure it is really simple, but i can't see it.
 
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If [itex]h(x) = 3x^2 - 3 [/tex]<br /> <br /> [tex]h'(x) = 6x[/tex]<br /> <br /> Critical point is at x = 0<br /> <br /> Decreasing at x < 0, Increasing at x > 0.[/itex]
 
how did they get x < -1 implies |x| = -x and |x + 1| = -x - 1?

What does [itex]|x|[/itex] mean?

As whozum noted, your derivative is wrong for the first question.
 
There has to be some typo, if [tex]h(x) = 3x^2-3[/tex], then [tex]h'(x)=6x[/tex]
The [tex]3(x+1)(x-1)[/tex] is just a factored form of h(x).

EDIT: It seems the other posters already cleared up this question. Fix your derivative.
 
Jameson said:
There has to be some typo, if [tex]h(x) = 3x^2-3[/tex], then [tex]h'(x)=6x[/tex]
The [tex]3(x+1)(x-1)[/tex] is just a factored form of h(x).

EDIT: It seems the other posters already cleared up this question. Fix your derivative.


sorry, I meant h'x = 3x^2 - 3.

and factor that to get 3(x+1)(x-1)...how did all the cases come up?
 
well, when is x-1 positive? When is x+1 positive? From that, can you tell when their product is positive?
 
Or simply plot that parabola.You know where its vertex is and you can find its intercepts with the Ox axis...


Daniel.
 

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