(B) Sketch the graph of f (x) = 1 − |x|; note that f (–1) = f (1) = 0 and that f
is continuous on [−1,1]. Only (B) holds.(C) Since f ′(x) = 6x^2 − 3, therefore ; also, f(x), or 2x^3 − 3x,
equals −1, by observation, for x = 1. So h′(−1) or (when x = 1)
equals (D)(B) Since (D) The given limit is the derivative of g(x) at x = 0.(B) The tangent line appears to contain (3,−2.6) and (4,−1.8).(D) f ′(x) is least at the point of inflection of the curve, at about 0.7.(C)(B) By calculator,