Mark M. answered 11/09/17
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Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.
f(x) = x2 / (x2 + 3)
f'(x) = [(x2+3)(2x) - x2(2x)] / (x2+3)2
= 6x/(x2+3)2
f'(x) = 0 when 6x = 0. So, x = 0
When x < 0, f'(x) < 0. So, f is decreasing on (-∞,0)
When x > 0, f'(x) > 0. So, f is increasing on (0,∞)
Therefore, local min at (0, f(0)) = (0, 0) and there is no local max.
f"(x) = [(x2+3)2(6) - 6x(2)(x2+3)(2x)] / (x2+3)4
= (x2+3)[6(x2+3) - 24x2] / (x2+3)4
= (18 - 18x2) / (x2 + 3)3
f"(x) = 0 when 18 - 18x2 = 0
x = ±1
When x < -1, f"(x) < 0. So, f is concave down on (-∞, -1).
When -1 < x < 1, f"(x) > 0, So, f is concave up on (-1,1).
When x > 1, f"(x) < 0. So, f is concave down on (1,∞).
Inflection points: (-1, f(-1)) = (-1, 1/4)
(1, f(1)) = (1, 1/4)