What are the points of inflection of $f (x) = \frac{x^{2}}{x^{2} + 49}$ $f(x)=x^2 / (x^2 + 49)$ ?

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Answer 1 · 2016-10-26T20:02:43

The points of inflection are $(\frac{7}{\sqrt{3}}, \frac{1}{4})$ $(7/sqrt3,1/4)$ and $(- \frac{7}{\sqrt{3}}, \frac{1}{4})$ $(-7/sqrt3,1/4)$

The points of inflections are when $f''(x)=0$

Let start by calculating the derivatves, we use $f'(u/v)=(u'v-uv')/v^2$

So $f'(x)=(2x(x^2+49)-x^2(2x))/(x^2+49)^2=(98x)/(x^2+49)^2$

And
$f''(x)=(98(x^2+49)^2-98x(x^2+49)*2*2x)/((x^2+49)^4)$

$=(98(x^2+49)((x^2+49)-4x^2))/((x^2+49)^4)$
$=(98(49-3x^2))/((x^2+49)^3$
so $f''(x)=0$ when $49-3x^2=0$

that is $x=+-7/sqrt3$
And $y=(49/3)/(49/3+49)=1/4$
So the points are $(7/sqrt3,1/4)$ and $(-7/sqrt3,1/4)$

What are the points of inflection of f(x)=x^2 / (x^2 + 49) f(x)=x2x2+49f(x)=x^2 / (x^2 + 49) ?