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{\bf Question}

Let $f_a(x)=x^4-3ax^2+x+2a^2$. Show that fixed points for $f_a$
are created as $a$ increases through zero, but this is {\underline
not} a saddle-node bifurcation.  Which conditions in Theorem A
fail?
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{\bf Answer}

$f_a(x)=x$ where $x^4-3ax^2+2a^2=0$ i.e. $(x^2-a)(x^2-2a)=0.$  No
solutions for $a>0$; solutions $\pm a, \pm \sqrt 2a$ for $a>0$. We
have $f_0'(0)=1$ (OK for saddle-node) but
$\left.\ds\frac{\partial}{\partial
a}f_a(0)\right|_{a=0}=\left.4a\right|_{a=0}=0$: second condition
fails.
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