# What are the points of inflection of f(x)=12x^3 + 3x^2 + 42x ?

Jan 17, 2016

$x = - \frac{1}{12}$

#### Explanation:

Points of inflection occur when the second derivative a function changes sign (from positive to negative, or vice versa). This correlates to when the concavity of the function shifts.

First, find the second derivative of the function.

$f \left(x\right) = 12 {x}^{3} + 3 {x}^{2} + 42 x$

$f ' \left(x\right) = 36 {x}^{2} + 6 x + 42$

$f ' ' \left(x\right) = 72 x + 6$

The sign of $f ' ' \left(x\right)$ could shift when $f ' ' \left(x\right) = 0$, so set $72 x + 6 = 0$.

$72 x + 6 = 0$

$x = - \frac{6}{72}$

$x = - \frac{1}{12}$

Analyze the sign of $f ' ' \left(x\right)$ around the possible point of inflection $x = - \frac{1}{12}$.

When $m a t h b f \left(x < - \frac{1}{12}\right)$:

$f ' ' \left(- 1\right) = - 72 + 6 = - 66$

This is $< 0$.

When $m a t h b f \left(x > - \frac{1}{12}\right)$:

$f ' ' \left(0\right) = 6$

This is $> 0$.

Thus, the sign of $f ' ' \left(x\right)$ does change around the point when $x = - \frac{1}{12}$.

We can check a graph of $f \left(x\right)$:

graph{12x^3+3x^2+42x [-2.5, 2.5, -200, 200]}

The concavity does seem to shift very close to $x = 0$.