$\displaystyle

f(x)=\sqrt{1+x^3}$

find the maximum value of |f''(x)| on the given Interval of $\displaystyle [0,2]$

calculated $\displaystyle f''(x)$ to be

$\displaystyle

f''(x) = \frac{3x(x^3+4)}{4(x^3+1)^{3/2}}$

thot setting this to zero would give where the maximum value is

but the answer is $\displaystyle f''|(\sqrt[3]{-10+\sqrt{108}})|=1.47$

graphing this can see that the top of the graph is 1.47 but don't see how this was derived.