It makes sense to analyze the LHS as a rational function. Simplify it to
f(x) = ((1+2a)x^4 + 5ax^2+ 4a) / (1+x^2)^2
The denominator is never 0 so there are no vertical asymptotes.
The numerator and denominator have the same degree so a graph would have a horizontal asymptote x = 1+2a.
f(0)=4a.
We really only care about the zeros of the numerator which is a biquadratic. z=x^2.
Has discriminant -7a^2-16a from which -16/7 <= a <= 0.
But also z cannot be negative
(-5a-sqrt(-7a^2-16a))/(4a+2))>=0 from which a=0
or
(-5a+sqrt(-7a^2-16a))/(4a+2))>=0 from which -0.5<a<=0
So the solution is a∈(-0.5,0]
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Posted by Jer
on 2024-08-30 10:45:00 |
Solution