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

##### 1 Answer
Oct 26, 2016

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

#### Explanation:

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

Let start by calculating the derivatves, we use $f ' \left(\frac{u}{v}\right) = \frac{u ' v - u v '}{v} ^ 2$

So $f ' \left(x\right) = \frac{2 x \left({x}^{2} + 49\right) - {x}^{2} \left(2 x\right)}{{x}^{2} + 49} ^ 2 = \frac{98 x}{{x}^{2} + 49} ^ 2$

And
$f ' ' \left(x\right) = \frac{98 {\left({x}^{2} + 49\right)}^{2} - 98 x \left({x}^{2} + 49\right) \cdot 2 \cdot 2 x}{{\left({x}^{2} + 49\right)}^{4}}$

$= \frac{98 \left({x}^{2} + 49\right) \left(\left({x}^{2} + 49\right) - 4 {x}^{2}\right)}{{\left({x}^{2} + 49\right)}^{4}}$
=(98(49-3x^2))/((x^2+49)^3
so $f ' ' \left(x\right) = 0$ when $49 - 3 {x}^{2} = 0$

that is $x = \pm \frac{7}{\sqrt{3}}$
And $y = \frac{\frac{49}{3}}{\frac{49}{3} + 49} = \frac{1}{4}$
So the points are $\left(\frac{7}{\sqrt{3}} , \frac{1}{4}\right)$ and $\left(- \frac{7}{\sqrt{3}} , \frac{1}{4}\right)$