Find the range of this function

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The discussion revolves around determining the range of the function g, particularly whether it includes the value 0. Participants clarify that x=0 is not an asymptote, and the confusion stems from the function intersecting the y-axis at (0,0). It is noted that the function can cross its horizontal asymptote, which is likely y=0, not x=0. The behavior of the function around the critical points x=1 and x=-1 is analyzed to understand the sign of g in different intervals. Ultimately, the consensus is that further examination of the function's behavior is necessary to accurately define its range.
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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.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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