
Matthew S. answered 03/07/20
PhD in Mathematics with extensive experience teaching Calculus
a) Fundamental Thm of Calculus tells you that F'(x) = f(x). Therefore set f(x) = 0 to find critical points of F: (x2-1)/x is 0 when x = 1.
b) f is negative on (0, 1), so F is decreasing there
f is positive on (1, ∞) so F is increasing on (1, ∞)
c) For absolute min and max, evaluate F at critical point (there is only one of those) and endpoints of domain. Compare this finite set of function values to find the biggest and the smallest)
F(1/2) = t2/2 - ln(t) evaluated at 1 (lower limit) and 1/2 (upper limit)
= (1/8 - ln(1/2)) - (1/2 - ln(1)) = -3/8 - (-ln(2)) ≈ 0.318
F(1) = 0 (upper and lower limits of integration are the same, so result is 0)
F(3) = 32/2 - ln(3) - (1/2 - ln(1)) = 9/2 - ln(3) - 1/2 = 4 - ln(3) ≈ 2.901
Therefore 0 = F(1) is the absolute min
2.901 ≈ F(3) is the absolute max