Find the second derivative, f''(x):
f'(x)=(x-3)(x-1)d/dx(x-2)+(x-3)(d/dx(x-1))(x-2)+(d/dx(x-3))(x-1)(x-2)
f'(x)=(x-3)(x-1)+(x-3)(x-2)+(x-1)(x-2)
f''(x)=(x-3)(d/dx(x-1))+(x-1)(d/dx(x-3))+(x-3)((d/dx(x-2))+(x-2)(d/dx(x-3))+(x-1)(d/dx(x-2))+(x-2)((d/dx(x-1))
f''(x)=(x-3)+(x-1)+(x-3)+(x-2)+(x-1)+(x-2)
f''(x)=2(x-3)+2(x-1)+2(x-2)
Set f''(x)=0 and solve for x:
2(x-3)+2(x-1)+2(x-2)=0
2x-6+2x-2+2x-4=0
6x-12=0
6x=12
x=12/6=2
The domain of f(x) is (-∞,∞), as all polynomials are continuous. Let's break up the domain of f(x) around the value of x we've found:
(-∞,2), (2,∞)
Now, we must determine whether f''(x) is positive or negative in each of these intervals. If f''(x)>0 in an interval, f(x) is concave on that interval. If f''(x)<0 on an interval, f(x) is convex on that interval.
(-∞,2):
f''(0)=2(-3)+2(-1)+2(-2)<0
f(x) is convex on the interval (-∞,2)
(2,∞):
f''(3)=2(0)+2(2)+2>0
f(x) is concave on the interval (2, ∞)
