Science:Math Exam Resources/Courses/MATH102/December 2019/Question 08(a)/Solution 1

The zeros of the functions are given by N = 0, 1, 2. The first derivative of the function is

$f'(N)=-3{\frac {N^{2}}{2}}+3N-1$ thus the critical points are given by

$N_{c}=1\pm {\frac {1}{\sqrt {3}}}$

Since the derivative is a downward quadratic it's easy to observe that

$N_{1}=1-{\frac {1}{\sqrt {3}}}$ is minimum , while $N_{2}=1+{\frac {1}{\sqrt {3}}}$ is maximum. The second derivative is given by $f''(N)=3(1-N)$ and indeed since it's a linear function we must have a point of inflection at N = 1.