Help with using the first derivative

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Homework Help Overview

The problem involves using the first derivative to determine where the graph of the function y = x/(x^2+1) is rising. The discussion centers around the analysis of the derivative and its implications for the behavior of the graph.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to analyze the derivative they calculated, seeking clarification on how to determine where it is positive. Some participants suggest focusing on the numerator of the derivative to identify intervals of positivity and negativity.

Discussion Status

Participants are actively discussing the implications of the derivative's factors and their signs. There is an ongoing exploration of the conditions under which the derivative is positive, with some guidance provided on factoring and analyzing the critical points.

Contextual Notes

There is some confusion regarding the correct interpretation of the factored form of the numerator and its implications for the sign of the derivative. Participants are also clarifying the intervals where the derivative is positive or negative.

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Homework Statement



Use the first derivative to determine where the graph of y = x/(x^2+1) is rising.

Homework Equations





The Attempt at a Solution



Ive figured the derivative to be (1-x^2) / (x^2+1)^2 and I know that the derivative > 0 will tell me where the graph is rising. I am just not sure how to figure that out. Do I need to simplify my derivative a bit more to make it work?
 
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The denominator will always be positive, so all you need to do is determine where the numerator is positive, and where negative. Factor 1 - x^2 and see where it is zero, and where positive, and where negative.
 
"Factor 1 - x^2 and see where it is zero, and where positive, and where negative."

Im not sure I know what you mean here. Factored form it is (x-1)(x+1)

Its 0 when x = +-1 , positive for 0<x<1 negative for (-infin,0) (1,infin)?
 
i think you may have missed the effect of one of the factors...

you could say (1-x^2) is +ve:
when
1-x^2 >0
implying
x^2 < 1
 
meeklobraca said:
"Factor 1 - x^2 and see where it is zero, and where positive, and where negative."

Im not sure I know what you mean here. Factored form it is (x-1)(x+1)

Its 0 when x = +-1 , positive for 0<x<1 negative for (-infin,0) (1,infin)?
"Factored form" is NOT (x- 1)(x+ 1), it is (1- x)(1+ x).

And the graph of y= (1-x)(1+x)= 1- x2, is a parabola opening downward and so y is positive for x between -1 and 1. I don't where you got the "0" in "0< x< 1". Was that a typo?
 
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