Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 41

You need to determine the monotony of the function, hence, you need to find the intervals where f'(x)>0 or f'(x)<0.
You need to find the derivative of the function, using the product rule:
f'(x) = (x^(1/3))'(x+4) + (x^(1/3))(x+4)'
f'(x) = (x+4)/(3x^(2/3)) + (x^(1/3))
You need to solve for x the equation f'(x) =0:
(x+4)/(3x^(2/3)) + (x^(1/3)) = 0
x + 4 + (3x^(2/3))(x^(1/3)) = 0
x + 4 + 3x = 0 => 4x + 4 = 0 => x = -1
Hence, f'(x)<0 and the function decreases for x in (-oo,-1) and f'(x)>0, the function increases for x in (-1,oo) .
b) Since for x = -1, f'(-1) = 0 and considering the monotony of the function, yields that the function has minimum point at x = -1.