Science:Math Exam Resources/Courses/MATH102/December 2015/Question 13/Solution 1

Note that all local minimums occur at critical points, so that critical points are the candidates of the point which achieves absolute minimum.

Therefore, we first find the critical points. Since $f^{\prime }(x)=(-x^3+3x+1{)}^{\prime }=-3x^2+3=-3(x-1)(x+1)$, the critical points are $x=\pm 1$. (i.e, $f^{\prime }(\pm 1)=0$)

Since $x=1$ is the only point in the given interval $[0,3]$, we compare the function value at this point with the ones at the boundary points, to find the absolute minimum;

$f(0)=1$, $f(1)=3$, and $f(3)=-27+9+1=-17$.

This implies that we have the absolute minimum $f(3)=-17$.