# Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (f)/Solution 2

(Alternative solution) From the solution 1, we know that the only critical point is ${\textstyle 2}$. Now let’s calculate the second derivative
${\displaystyle f''(x)=(f'(x))'=({\frac {2-x}{x^{3}}})'={\frac {(2-x)'x^{3}-(x^{3})'(2-x)}{x^{6}}}={\frac {2x-6}{x^{4}}}.}$
At ${\textstyle x=2}$, ${\textstyle f''(2)={\frac {2\times 2-6}{2^{4}}}=-{\frac {1}{8}}<0}$, so by the second derivative test in the Hint 2, it attains local maximum at this point.
In all, it has no local minimum, but has a local maximum at ${\textstyle x=2}$ and ${\textstyle f(2)={\frac {2-1}{x^{2}}}={\frac {1}{4}}}$.