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Quiz #2-Joe March 7, 2014 Name: Math 151 Section Row: This quiz is due by 11:30 on March 7. Do you work either at the bottom of this page, the back of this paper or on another sheet of paper. Clearly label your work and your answers. 1. For y = x2 find y ′′ (x). 3x + 1 y′ = 3x2 + 2x (3x + 1)(2x) − x2 (3) = 2 (3x + 1) (3x + 1)2 y ′′ = (3x + 1)2 (6x + 2) − (3x2 + 2x) ∗ 2(3x + 1)(3) (3x + 1)[(3x + 1)(6x + 2) − 6(3x2 + 2x)] = 4 (3x + 1) (3x + 1)4 y ′′ = 2 18x2 + 12x + 2 − 18x2 − 12x = 3 (3x + 1) (3x + 1)3 2. For f (x) = 1 find f (3) (x). (7x + 1)2 f (x) = (7x + 1)−2 f ′ (x) = −2(7x + 1)−3 ∗ 7 f ′′ (x) = 6(7x + 1)−4 ∗ 72 f (3) (x) = −24(7x + 1)−5 ∗ 73 or f (3) (x) = −24 ∗ 73 (7x + 1)5 3. If x = 3t + 5 and y = t3 + 2t2 − 4. Compute dy dx (11,12) solve x = 3t + 5 = 11 for t to find the value of t that give the point. You get t = 2. 3t2 + 4t dy = dx 3 dy dy 20 = = dx (11,12) dx t=2 3 4. Find the values of t where the curve will have a horiozntal tangent line. x = t2 − 25 y = (t2 + 4t)2 dx dy = 2t = 2(t2 + 4t) ∗ (2t + 4). dt dt you want dy/dt = 0 and dx/dt = 6 0. dy = 2(t2 + 4t) ∗ (2t + 4) = 0 dt 2t(t + 4) ∗ 2(t + 2) = 0 This gives t = 0, t = −4, and t = −2. since t = 0 makes dx/dt = 0 it is not one of the answers. Answer: t = −4, and t = −2