(A) We have $f(x) = x^3 - 3x + 3$.
Taking the derivative,we get $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$.
Setting $f'(x) = 0$,we find critical points at $x = 1$ and $x = -1$.
To determine the nature of these points,we use the first derivative test:
For $x = 1$:
- For values close to $1$ and to the right,$f'(x) > 0$.
- For values close to $1$ and to the left,$f'(x) < 0$.
Since the sign changes from negative to positive,$x = 1$ is a point of local minima,and the local minimum value is $f(1) = 1 - 3 + 3 = 1$.
For $x = -1$:
- For values close to $-1$ and to the left,$f'(x) > 0$.
- For values close to $-1$ and to the right,$f'(x) < 0$.
Since the sign changes from positive to negative,$x = -1$ is a point of local maxima,and the local maximum value is $f(-1) = -1 + 3 + 3 = 5$.