Question

what is the largest value of curvature of the f(x)=(x)^3/3

Answer 1

http://en.wikipedia.org/wiki/Curvature#Curvature_of_a_graph

curvature is defined as:
\[k(x) = \frac{f''(x)}{(1+f'(x)^{2})^{3/2}}\]
to find largest value or maximize curvature, set derivative equal to zero

\[k'(x) = 0\]
first find f'(x) and f''(x) then find k'(x)

Answer 2

does that help? or are you having trouble taking the derivatives

Answer 3

@marsss that is completely irrelevant

Answer 4

@dumbcow *curvature for z = f(x) functions is defined as that :).

Answer 5

@ee2oo , im going to respond here
the denominator doesn't matter since it equals 0 .... just multiply both sides by denominator

Answer 6

@RONNCC , haha thx for correction

Answer 7

also to confirm, i think you got k'(x) right
\[\rightarrow k'(x) = \frac{2(1+x^{4})^{3/2} - 12x^{4} \sqrt{1+x^{4}}}{(1+x^{4})^{3}}\]

Answer 8

\[2(1+x^{4})^{3/2} - 12x^{4} \sqrt{1+x^{4}} = 0\]
\[\sqrt{1+x^{4}} [2(1+x^{4}) -12x^{4}] = 0\]
\[2 -10x^{4} = 0\]
\[x^{4} = \frac{1}{5}\]

Answer 9

@RONNCC you are an idiot :)