Find all inflection points of the following Hill function:
$\displaystyle f(x)= \frac{Ax^3}{(a^3 + x^3)}$
assuming that $a > 0$.
How do I approach this question?
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$$ f(x)=\frac{Ax^3}{a^3+x^3}=\frac{Aa^3+Ax^3-Aa^3}{a^3+x^3}=A-\frac{A a^3}{a^3+x^3} $$ $$ f'(x)=-A a^3 \,\frac{-3x^2}{(a^3+x^3)^2} $$ $$ f''(x)=-A a^3 \, \left( \frac{-6x}{(a^3+x^3)^2}-3x^2\frac{-6x^2}{(a^3+x^3)^3}\right)= \\ =-A a^3 \left( \frac{-6x(a^3+x^3)+6x 3x^3}{(a^3+x^3)^3}\right)=A a^3 \frac{6x(a^3-2x^3)}{(a^3+x^3)^3} $$ Then the possible (in fact, they are) points of inflection are $x=0$ and $x=\frac{a}{2^{1/3}}$.
Pocho la pantera
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Why is there a -3 in front of the x^2? and why are you subtracting the equation from A in the first line? – Lisa Paima Sep 30 '13 at 00:38
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i will edit to clarify it – Pocho la pantera Sep 30 '13 at 00:41
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Ready. There was a little mistake in the old calculation. I believe it is ok now. – Pocho la pantera Sep 30 '13 at 00:51
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Why is Aa^ 3 and not Ax^ 3? and also in the first line wouldn't +Aa^3 cancel out with -Aa^3 leaving us with Ax^3 as the numerator. – Lisa Paima Sep 30 '13 at 00:54
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I add and substract $A a^3$ to simplify the expression. – Pocho la pantera Sep 30 '13 at 00:57
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I'm sure you can verify that the first line is ok. The right hand side is easier to derivate. – Pocho la pantera Sep 30 '13 at 00:59