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,

$g(x)=6x|x|=\begin{cases}6x^2&\text{for }x\ge0\\-6x^2&\text{for }x<0\end{cases}$

Differentiating once gives

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

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

$g'(x)=\begin{cases}12x&\text{for }x\ge0\\-12x&\text{for }x<0\end{cases}$

Differentiating again, we find

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

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 $g''(x)<0$ when $x<0$, 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.)

Answer 2

The inflection point will be "$(x, g(x))= (0,0)$"

Given:

• $g(x) = 6x|x|$

now,

• $g'(x) = 6.\frac{d}{dx} (x|x|)$

As we know the product rule:

→ $\frac{d}{dx} (uv) = u.v'+v.u'$

then,

= $6[x.\frac{d}{dx} (|x|)+ |x|.\frac{d}{dx} (x)]$

= $6[x.\frac{|x|}{x}+|x|.x ]$

= $6[|x|+|x|.x]$

and,

→ $g''(x) = [6.\frac{d}{dx} {|x|+x.|x|}]$

            $= 6[\frac{d}{dx}(|x|)+\frac{d}{dx} (x.|x|) ]$

            $= 6[\frac{|x|}{x} +|x|+x|x|]$

            $= 6 |x| [\frac{1+x+x^2}{x} ]$

$g''(x)$ is undefined at x=0 because where denominator is "0".

At x = 0,

• $g(x)=6(0) |0|=0$

Thus the answer above is right.

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