So, the points of inflection are at x = ± 1.
f”(x) = 0 ⇒ x2 = 1 ⇒ x = ± 1
So, the intervals are (-∞, -1), (-1, 1), (1, ∞)
When x ∈ (-∞, -1), say x = -2
f”(x) = 12(-2)2 – 12 = 48 – 12 = 36 > 0
⇒ f(x) is concave upward in (-∞, -1) when x ∈ (-1, 1) say x = 0.
f”(x) = 0 – 12 = -12 < 0
⇒ f(x) convex upward.
⇒ x = -1 is a point of inflection. Again when x ∈ (-1, 1), the curve is convex upward and when x ∈ (1, ∞) say x = 2.
f”(x) = (1, ∞) say x = 2.
f”(x) = 12 (4) – 12 > 0
⇒ the curve is concave upward.
⇒ x = 1 is a point of inflection.
Thus x = ± 1 are the points of inflection.
At x = 1, f(x) = 1 – 6 = -5
At x = -1, f(x) = (-1) – 6(-1) = 1 – 6 = -5
The curve concave upward in (-∞, -1) u (1, ∞) and convex upward in (-1, 1) and the points of inflection are (1, -5) and (-1, -5).