I am not English speaking and also highlight is yellow in solution…
The graph in the fogure shows an even function $f(x) = \dfrac{p(x)}{q(x)}$ where $p(x)$ and $q(x)$ are rational quadratic polynomials. Give possible formulas for $p(x)$ and $q(x)$.
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Ѕᴀᴀᴅ
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It is better to insert the image rather than a link to the image, as well as providing the relevant parts of the text, the statement and the beginning of the proof, rather than two links. – Hetebrij Apr 13 '18 at 11:46
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From the first picture we are meant to see that $f(0)=0$, and also that the $x$-axis is a tangent, so that $f'(0)=0$. That's what a "second-order zero" means. That means the same is true for $p(x)$, so that $x^2$ must divide $p(x)$.
The constant $\chi$ can't be $0$ or else an $x$ on the denominator would cancel out one of the $x$s on top, and leave us with just a simple zero, not a second-order one.
ancient mathematician
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It means degree of p is 2 (or x^2 is not zero), am i right about second order zero?the figure shows it is second order zero????
Thanks!
– Mustafa Asghari Apr 13 '18 at 13:14 -
No, the solution assumes the degree of $p$ is $2$. Looking at the graph of $f$ one can "see" that it passes thro $0$ and is flat there, so that $x^2$ must divide $p$. – ancient mathematician Apr 13 '18 at 16:05
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d(P(x)) is 0 is second order zero???derivative of p... ?? what is 1st order zero or 3rd for example? – Mustafa Asghari Apr 14 '18 at 08:27
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If $F(x)=(x-a)^n G(x)$ with $G(a)\not=0$ then $F$ has a zero of order $n$ at the point $a$. – ancient mathematician Apr 14 '18 at 12:26