Find the range of this function

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Homework Help Overview

The discussion revolves around determining the range of a function g, defined for x not equal to ±1. Participants are exploring the implications of asymptotes and function behavior near critical points.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are questioning whether the range includes the value 0, given its relationship to the function's asymptote and intersection with the y-axis. There is also discussion about the behavior of the function near the points where it is undefined.

Discussion Status

The conversation is ongoing, with some participants providing clarifications about asymptotes and suggesting further examination of the function's behavior in different intervals. No consensus has been reached yet.

Contextual Notes

There is confusion regarding the definition of asymptotes, particularly whether y = 0 or x = 0 is the relevant asymptote. The problem context includes the function's undefined points at x = ±1.

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Homework Statement



A function g is defined by
MSP638619fcf3dddfha5i5a00005bd712931f7b0gbb.gif
, x=/= +-1 . Determine the range of of g.

Homework Equations


The Attempt at a Solution



Is it {y:y∈R} or {y:y∈R, y=/=0}? Is ''0'' included? Because x=0 is the asymptote of the function but at the same time, function g also intersect the y-axis at (0,0). So i am confused on whether the answer should include 0 or not... Can anyone enlighten me? Thanks.
 
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x=0 is not an a asymptote.

You already know that x=1 and x=-1 cause the function to divide by zero - what happens immediately on either side of x=1 and -1?
 
I'm guessing that the OP should have meant "y = 0" as an asymptote, not "x = 0."

A function can intersect its horizontal asymptote in the "middle" of the graph. Just not at the "ends" -- when x approaches infinity and/or negative infinity.
 
Right, you know that when 0<x<1 this means that 1-x^2>0 and therefore, g will be positive, for 0>x>-1, g(x) will be negative. Using this, examine when x>1 and x<-1.
 

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