William C. answered 11/08/23
Experienced Tutor Specializing in Chemistry, Math, and Physics
f(x) = 6x2ln(x)
(a)
f’(x)= 12xln(x) + 6x2/x = 12xln(x) + 6x = 6x(2ln(x) +1)
f’(x)= 0 when 2ln(x) + 1 = 0 which means
f(e–1/2) = –3/e
f(x) and f’(x) are both undefined at x ≤ 0
(b)
(f), (g), and (h)
f’’(x) = 12ln(x) +12x/x +6 = 12ln(x)+18 = 6(2ln(x) + 3)
f(x) has an inflection point at x = e–3/2
On the interval (0, e–3/2) f''(x) < 0 so f(x) is concave down
On the interval (e–3/2,∞) f''(x) > 0 so f(x) is concave up