Question

Find the inflection point of the function. (hint: g''(0) does not exist.) g(x) = 6x|x|

Answer 1

By definition of absolute value,

$g(x)=6x|x|=\begin{cases}6x^2&\text{for }x\ge0\\-6x^2&\text{for }x$

Differentiating once gives

$g'(x)=\begin{cases}12x&\text{for }x>0\\?&\text{for }x=0\\-12x&\text{for }x$

As $x\to0$ from either side, we find that $g'(x)\to0$, so really

$g'(x)=\begin{cases}12x&\text{for }x\ge0\\-12x&\text{for }x$

Differentiating again, we find

$g''(x)=\begin{cases}12&\text{for }x>0\\?&\text{for }x=0\\-12&\text{for }x$

but this time, the limit does not exist as $x\to0$ from either side, so $g''(0)$ is undefined. However, we see that $g''(x)>0$ when $x>0$, and $g''(x)$ when $x$, and we know that $g(x)$ is continuous at $x=0$. This means the concavity must change at $x=0$, so (0, 0) is the inflection point. (The takeaway is that inflection points *can* occur when the second derivative is undefined, but not always.)