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A family of functions $\{f_{a,b} \mid a, b \in \mathbb{R}\}$ is given thru $$f_{a,b}(x)=\frac{x^4+ax^2+b}{x^2+1}.$$ It is asked to pick those functions such that their graph $\Gamma_{a,b}$ is tangent to the horizontal axis in two distinct points.

I have written the conditions of tangency and of passage thru a point $(x,0)$. The discussion is a bit long, but possible. I wonder if there is any "shortcut", i.e. some approach which leads to the solution more quickly.

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What tangency condition have you established? – Mark Bennet Apr 29 '12 at 17:33
The derivative of the function must vanish at two distinct points. – Siminore Apr 29 '12 at 17:58
Put $x^2:=u$ and construct tangencies $(u,0)$ with $u>0$. – Christian Blatter Apr 29 '12 at 17:59
You should have that $f$ and $f'$ are simultaneously zero. This should give you two polynomial equations, and the condition you need is that they have a common factor - if all else fails, you could use the division algorithm for polynomials. Note that $f$ is an even function - evidently if $x$ is a solution so is $-x$, so most of the time you only need to find one point - another comes automatically. Christian's hint uses this to simplify the calculations. There is a special case $x=0$ to consider. – Mark Bennet Apr 29 '12 at 18:17

The derivative is $(4x^3+2ax)(x^2+1)-(2x)(x^4+ax^2+b)$ divided by something harmless. The $x^4+ax^2+b$ part will have to be $0$, which means we will want $4x^3+2ax$ to be $0$. Short enough?

If we want one of the points of tangency to be at $x=0$, we get $b=0$, and want $4x^2+2a$ and $x^4+ax^2$ to be simultaneously $0$ at some $x\ne 0$. Is this possible?

If $x=0$ is not a point of tangency, we want $4x^2+2a=0$, $x^4+ax^2+b=0$. Should be easy to find out what $b$ must be in terms of $a$. (The double part is not much of an issue because of symmetry.)

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