Finding the maximum curvature

Question

I am trying to find the maximum curvature of $y=1/x$. I know to begin, I find k(x), which is: $k(x) = \frac {2}{x^3 (1+1/x^2)^{3/2}}$. But I'm confused as to where to go from here.

Answer 1

The maximum (or minimum) of any function should happen when $f'(x)=0$.

In this case $f(x) = k(x)$, so set the derivative of the curvature to zero $k'(x)=0$ and solve for $x$. The use this $x$ to get the value of the curvature.