50. [4/7 Points] DETAILS PREVIOUS ANSWERS LARHSCALC2 4.AP.011. MY NOTES ASK YOUR TEACHER PRACTICE AN Do not use technology. For 0 0. At these times, x(t) 0. At these times, a(t) > 0. At these times, a(t) 0, the particle is speeding up. Because v(5) 0, the particle is speeding up. Because v(5) = 0, the particle is neither speeding up nor slowing down.F( 4) = -0.5 (b) Find the x-coordinates of all the relative minimum values of F(x) on the interval [-3, 4]. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNB X = -3,3 X Justify your answer. There are no relative minimum values because F(x) does not change from positive to negative at any point. The relative minimum values occur where F(x) changes from negative to positive. The relative minimum values occur where f(x) changes from negative to positive. There are no relative minimum values because f(x) does not change from negative to positive at any point. The relative minimum values occur where f(x) changes from positive to negative. (c) Find the x-coordinates of the inflection points of F(x) on the interval [-3, 4]. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) X = 0 Justify your answer. There are no inflection points because f"(x) never changes sign. The inflection points occur where f"(x) changes sign. The inflection points occur where F"(x) changes sign. There are no inflection points because F"(x) never changes sign. X (d) Write the equation of the line tangent to F(x) at the point where x = 2. y ( x ) = Need Help? Read ItLet F ( x ) = / "f(t ) at . The graph of f on the interval [-3, 4] consists of two line segments and a semicircle, as shown in the graph. (a) Find F(0), F'(0), and F(4). F(0) = -5.64 X F'(0) = 3 F(4) = -0.5 (b) Find the x-coordinates of all the relative minimum values of F(x) on the interval [-3, 4]. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) X = -3,3 X Justify your answer. There are no relative minimum values because F(x) does not change from positive to negative at any point. The relative minimum values occur where F(x) changes from negative to positive. The relative minimum values occur where f(x) changes from negative to positive. There are no relative minimum values because f(x) does not change from negative to positive at any point. The relative minimum values occur where f(x) changes from positive to negative.Justify your answer. At these times, v(t) > 0. At these times, x(t) 0. At these times, a(t) > 0. At these times, a(t) 0, the particle is speeding up. Because v(5) 0, the particle is speeding up. Because v(5) = 0, the particle is neither speeding up nor slowing down. (d) Find the position of the particle at time t = 5. (Round your answer to two decimal places.) x (5 ) = Need Help? Read It34. [-75 Points] DETAILS LAKNSCALL4 4.AP.UIS. IVIT NOTES ASK TUUR ICALMER PRACTICE ANUINER Do not use technology. The graph of a continuous function fis shown. The three regions between the graph of f and the x-axis are marked A, B, and C, and have unsigned areas 6.5, 5, and 13.5, respectively. Let F(x) be an antiderivative of f that is differentiable on (0, 4) and with F(1) = 9. C A B (a) Find F(0) and F(4). F(0) = F ( 4 ) = (b) What is the minimum number of times F equals 5 on the interval [0, 4]? c) Find all intervals where F is increasing. (Enter your answer using interval notation.) Justify your answer. OF is increasing where f(x) is concave upward. OF is increasing where f(x) is increasing. OF is increasing where F(x) is concave upward. OF is increasing where F(x) > 0. O F is increasing where f(x) > 0. Need Help? Read ItOn a typical day, the snow on a mountain melts at a rate modeled by the function M(t ) = _ sin A snow maker adds snow at a rate modeled by the function S(t) = 0.006t2 - 0.08t + 0.81. Both M and S have units in inches per hour and t is measured in hours for 0 s t s 4. At t = 0, the mountain has 30 inches of snow. (Round your answers to two decimal places.) (a) How much snow will melt during the 4 hour period? Indicate units of measure. 2 inches (b) Write an expression for I(t), the total number of inches of snow at any time t. I(t) = 0.00273 - 0.0472 + 0.81t + 2 cos + 28 (c) Find the rate of change of the total amount of snow when t = 2. 0.34 X inches per hour (d) For 0 s t s 4, at what time t is the amount of snow a maximum? 4 x hours What is the maximum value? 31.88 x inches Justify your answers. Because I(t) is decreasing over [0, 4], the maximum is at t = 0. Because I(t) is concave upward over [0, 4], the maximum is at t = 0. Because I(t) is increasing over [0, 4], the maximum is at t = 4. Because I(t) is concave downward over [0, 4], the maximum is between t = 0 and t = 4. Because I(t) is concave upward over [0, 4], the maximum is at t = 4. X Need Help? Read It Culpait A