For the curve ,

MATHEMATICS · Single correct · AP · Calculus BC · Parametric, Polar, and Vector-Valued Functions · Parametric, Polar, and Vector-Valued Functions

Problem

For the curve x=t2+1x = t^{2}+1, y=t3−3ty = t^{3}-3t, consider the open interval (−3,3)(-3,3) of the parameter tt. On which part of this interval is the curve concave upward? (a) (−3,0)(-3,0) only (b) (0,3)(0,3) only (c) (−3,0) ∪ (0,3)(-3,0)\,\cup\,(0,3) (d) Nowhere in (−3,3)(-3,3)

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