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LammettHash
By definition of absolute value,

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

Differentiating once gives

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

As [tex]x\to0[/tex] from either side, we find that [tex]g'(x)\to0[/tex], so really

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

Differentiating again, we find

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

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

The inflection point will be "[tex](x, g(x))= (0,0)[/tex]"

Given:

  • [tex]g(x) = 6x|x|[/tex]

now,

  • [tex]g'(x) = 6.\frac{d}{dx} (x|x|)[/tex]

As we know the product rule:

→ [tex]\frac{d}{dx} (uv) = u.v'+v.u'[/tex]

then,

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

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

= [tex]6[|x|+|x|.x][/tex]

and,

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

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

            [tex]= 6[\frac{|x|}{x} +|x|+x|x|][/tex]

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

[tex]g''(x)[/tex] is undefined at x=0 because where denominator is "0".

At x = 0,

  • [tex]g(x)=6(0) |0|=0[/tex]

Thus the answer above is right.

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