# What are the points of inflection, if any, of f(x)=3 x^5 + 4 x^3 - x^2-x ?

May 11, 2018

There is one point of inflection at $x \approx 0.23$.

#### Explanation:

Possible points of inflection exist whenever $f ' ' \left(x\right) = 0$. We then test intermediate values to determine if they are truly points of inflection.

$f \left(x\right) = 3 {x}^{5} + 4 {x}^{3} - {x}^{2} - x$
$f ' \left(x\right) = 15 {x}^{4} + 12 {x}^{2} - 2 x - 1$
$f ' ' \left(x\right) = 60 {x}^{3} + 24 x - 2$

Letting $f ' ' \left(x\right) = 0$ gives:

$60 {x}^{3} + 24 x - 2 = 0$
$30 {x}^{3} + 12 x - 1 = 0$

This is a cubic equation which is by no means easy to solve by factoring or guessing and checking. To solve it, you'll need to utilize the cubic formula. It's like the quadratic formula except it works for cubics and it is significantly longer. Unless you enter some cruel mathematics competition where competitors are expected to be able to solve cubics like these, I wouldn't bother memorizing it.

The cubic formula reveals a single real root:

$x = \frac{1}{30} \left(- 4 + \sqrt{386 - 30 \sqrt{161}} + \sqrt{386 + 30 \sqrt{161}}\right)$

Or $x \approx 0.23$

We now need to check whether there is actually a change in concavity at this point. We do this by plugging in some $f ' ' \left(a\right) , a < 0.23$ and some $f ' ' \left(b\right) , b > 0.23$. If the signs change from positive to negative or vice-versa, there is a point of inflection.

$f ' ' \left(0\right) = - 1$
$f ' ' \left(1\right) = 30 + 12 - 1 = 41$

The signs do change, so we have a point of inflection at $x \approx 0.23$.