# What are the points of inflection of f(x)=1/(5x^2+3) ?

Jun 6, 2018

No inflection points

#### Explanation:

Calculate the zeroes of $f ' \left(x\right)$. The points of inflection are these points where $f ' \left(x\right) = 0$ and also $f ' ' \left(x\right) = 0$.

$f ' \left(x\right) = - \frac{10 x}{5 {x}^{2} + 3} ^ 2$

$f ' \left(x\right) = 0 \Rightarrow - 10 x = 0 \Rightarrow x = 0$, a single possible point of inflection. Note that both $f$ and $f '$ are well-defined at this point.

Calculate $f ' ' \left(x\right)$ to determine the nature of the point:

$f ' ' \left(x\right) = \frac{- 10 {\left(5 {x}^{2} + 3\right)}^{2} + 10 x \left(2 \left(5 {x}^{2} + 3\right) \cdot 10 x\right)}{5 {x}^{2} + 3} ^ 4$
$= \frac{- 10 {\left(5 {x}^{2} + 3\right)}^{2} + 200 {x}^{2} \left(5 {x}^{2} + 3\right)}{5 {x}^{2} + 3} ^ 4$
$= \frac{- 10 \left(5 {x}^{2} + 3\right) + 200 {x}^{2}}{5 {x}^{2} + 3} ^ 3$

Therefore $f ' ' \left(0\right) = \frac{- 30 + 0}{3} ^ 3 = - \frac{30}{27} < 0$. So this point is a local maximum of $f$, not a point of inflection.

Compare with the graph of the function for a visual sanity check:
graph{1/(5x^2+3) [-2.5, 2.5, -1.25, 1.25]}