# What are the points of inflection, if any, of f(x)=3x^4 − 6x^3 + 4 ?

Jun 15, 2016

They occur at $x = 0$ and $x = 1$.

#### Explanation:

Points of inflection are where a function's concavity shift from concave upwards to concave downwards, or vice versa.

Since concave upwards functions occur when the second derivative is positive, and concave downwards correlates to the second derivative being negative, a point of inflection occurs at the switching point when the second derivative equals zero.

So, we need to set the second derivative equal to $0$ and check those points to make sure the second derivative goes from positive or negative or negative to positive, and doesn't just stay positive or negative and "bounce" off $0$.

The second derivative can be found through applying the power rule:

$f \left(x\right) = 3 {x}^{4} - 6 {x}^{3} + 4$

$f ' \left(x\right) = 12 {x}^{3} - 18 {x}^{2}$

$f ' ' \left(x\right) = 36 {x}^{2} - 36 x$

Points of inflection could occur when:

$36 {x}^{2} - 36 x = 0$

$36 x \left(x - 1\right) = 0$

So, when $x = 0$ or $x = 1$.

Test the sign of the second derivative around these points:

$f ' ' \left(- 1\right) = 36 \left(- 1\right) \left(- 1 - 1\right) = 72 > 0$

$f ' ' \left(0\right) = 0$

$f ' ' \left(\frac{1}{2}\right) = 36 \left(\frac{1}{2}\right) \left(\frac{1}{2} - 1\right) = - 9 < 0$

$f ' ' \left(1\right) = 0$

$f ' ' \left(2\right) = 36 \left(2\right) \left(2 - 1\right) = 72 > 0$

Around $x = 0$, the second derivative switches from positive to negative, so there is a point of inflection there. Similarly, since the sign of the second derivative also shifted around $x = 1$, it is also a point of inflection.