f ′(1) exists and is equal to 2.(B) Since |x − 2| = 2 − x if x < 2, the limit as .(A)Note that the distance covered in 6 seconds is , the area between
the velocity curve and the t-axis.(C) Acceleration is the slope of the velocity curve, .(D) Slopes are: 1 along y = x, −1 along y = −x, 0 along x = 0, and
undefined along y = 0.(A) Differentiating implicitly yields 2xyy′ + y^2 − 2y′ + 12y^2 y′ = 0. When y
= 1, x = 4. Substitute to find y′.(B)