Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 19

f(x) is concave up

First derivative,
f'(x) = 1*(x-4)^3+x*3*1*(x-4)^2
f'(x) = (x-4)^2(x-4+3x)
f'(x) = 4(x-4)^2(x-1)
There are only two critical points, x = 4 and x = 1
Second derivative,
f''(x) = 4(2*1*(x-4)*(x-1)+ (x-4)^2*1)
f''(x) = 4(x-4) (2x-2+x-4)
f''(x) = 4(x-4)(3x-6)
f''(x) = 12(x-4)(x-2)
For inflection points, f''(x) = 0.
Therefore, the inflection point is at x = 4, not at x = 1.
x<2
f"(x) >0, therefore f'(x) is increasing, which means f(x) is concave up at x<2.
2f"(x) < 0, therefore f'(x) is decreasing, which means f(x) is concave down at 2x>4
f"(x) > 0, therefore f'(x) is increasing, which means f(x) is concave up at x>4