Simplfing the second derivative of x + 3x^(-1)

[cmnalo]

g(x) = x + 3x^(-1)
g'(x) = 1 - 3x^(-2)
g"(x) = 6x^(-3) --> 6/x^3

Is this correct? Any guidence would be appreciated. I eventually need to find the inflection point.

 

[galactus]

What does the second derivative say about inflection pts?.

Check the increasing and decreasing intervals of f'(x).

Note the vertical asymptote at x=0.

 

[cmnalo]

So, assuming I have the correct second derivative f"(x)=6/x^3 than I would make that equal to zero to determine the points in the domain of f. f(x) is discontinous at x=0 correct? For each interval the concavity doesn't change correct? So there is no point of inflection. I'm barely understanding this and I not sure everything I'm saying even makes sense. Could you please offer me some clarification.

 

[galactus]

It is, indeed, discontinuous at x=0. There are no inflection pts.

\(\displaystyle \L\\\lim_{x\to\0^{+}}(x+\frac{3}{x})\;\ \rightarrow \;\ {+\infty}\)

\(\displaystyle \L\\\lim_{x\to\0^{-}}(x+\frac{3}{x})\;\ \rightarrow\;\ {-\infty}\)