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#1 – No Calc

#1 – No Calc. C. #2 – No Calc. Then use x=7 to solve for A , which is 3. So one of the fractions obtained is. C. #3 – No Calc. At t = 1 ,. B. #4 – No Calc. C. #5 – No Calc. when x=0, x=2, or x=-2 . - -2 + 0 - 2 +

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#1 – No Calc

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  1. #1 – No Calc C

  2. #2 – No Calc Then use x=7 to solve for A, which is 3. So one of the fractions obtained is C

  3. #3 – No Calc At t = 1, B

  4. #4 – No Calc C

  5. #5 – No Calc when x=0, x=2, or x=-2. - -2 + 0 - 2 + f(x)has a relative min when f’(x) changes from – to +. Therefore, f(x) has relative minima at x=2 and x=-2. D

  6. #6 – No Calc A

  7. #7 – No Calc The tangent line to the curve is horizontal when dy/dx = 0, so when t = ½ . D

  8. #8 – No Calc If the function has a horizontal tangent for some value of x, then y’=0 for that value of x. If the function has a point of inflection, then y’’=0. Plugging in to y’ gives us So b = -6. A

  9. #9 – No Calc Euler’s method is A

  10. #10 – No Calc The interval of convergence includes 0.5 but not 1.5. All Taylor polynomials are equal to the function at the center of the interval of convergence. I and III are true. D

  11. #11 – No Calc Substitutions give the indeterminate form 0/0. Using L’Hopital’s Rule: D

  12. #12 – No Calc I and III converge. E

  13. #13 – No Calc By implicit differentiation Solving for dy/dx, B

  14. #14 – No Calc The complete derivative with the correct Chain Rule inclusion is only found in choice E. Let u=g(x) and du=g’(x)dx. E

  15. #15 – No Calc Integrate the velocity function to find position: The velocity is positive until t=4. At that point in time it stops and then begins moving down. From the information in the problem s(4)=0. Find C. So A

  16. #16 – No Calc Using So E

  17. #17 – No Calc B

  18. #18 – No Calc Since so k=3. E

  19. #19 – No Calc The first term is ¼. The ratio is also ¼. The sum of a geometric series is where a is the first term in the series, and r is the ratio. Therefore, A

  20. #20 – No Calc B

  21. #21 – No Calc Or you should already know that the derivative of an exponential function includes the original function. E

  22. #22 – No Calc The graph is concave down when y’’<0. B

  23. #23 – No Calc Let y = the amount present at time t. We are given y(-4)=12 and y(0)=8. We need to find y(8). Since Therefore, C

  24. #24 – No Calc • Convergent: p-series with p>1 • Convergent: all terms less than p-series with p=2(which is known to be convergent) • Divergent: all terms greater than p-series with p=1(which is known to be divergent) C

  25. #25 – No Calc B

  26. #26 – No Calc The normal line has an opposite reciprocal slope, -2. So D

  27. #27 – No Calc Since the velocity, s’(t), is positive the graph must be increasing everywhere. Since the acceleration, s”(t), is positive the graph must be concave up everywhere. The only graph that matches these conditions is… C

  28. #28 – No Calc If on [0,1] and , then is a right-hand Riemann sum, and B

  29. #29 - Calculator Differentiate the position equations to find the velocity vector: Then differentiate to find the acceleration vector: Substitute t = 3, C

  30. #30 - Calculator The graph of f’(x) is increasing at x = 2, so f’’(2)>0. From the graph f’(2)=0. The derivative changes from – to + at x = 2, f(2) is a min. So f(2) < f(1), and f(2) < 0. Therefore f(2) < f’(2) < f’’(2). A

  31. #31 - Calculator • Since f’(x) changes from – to + at x = 0, x = 0 is a relative maximum. • f’(x) is always increasing so f(x) is concave up. • f(x) does not change concavity. • f(0) is not given and cannot be determined. • f’(x) is not even so f(x) cannot be odd. The derivative of an odd function must be even. A

  32. #32 - Calculator To maximize the catch set C’ equal to 0. indicating a maximum. B

  33. #33 – No Calculator • The function is decreasing to the right of the origin, so • As x approaches ∞ the graph flattens out so • As x approaches -∞ the graph flattens out so D

  34. #34 - Calculator Integrate the velocity vector to get the position vector: . When graphing this on the calculator the values for C and D will always produce a ray. E

  35. #35 - Calculator The acceleration becomes negative when the position function changes concavity from concave up to concave down. This corresponds to C. C

  36. #36 - Calculator E The derivative changes from negative to positive at x = 0.618 indicating a relative minimum. After the maximum at x = 1.623, the function decreases again making x = 5 a possible absolute minimum. The negative area is much greater than the positive area. This means the function lost a lot more than it gained from the last minimum. Therefore, x = 5 is the absolute minimum.

  37. #37 - Calculator The derivative can be approximated using two points near x = 4, 3.99900 and 4.00100 for instance: D

  38. #38 - Calculator A function can have only one horizontal asymptote. If it has a horizontal asymptote the value must be approached as x approaches infinity. B

  39. #39 - Calculator Graph the functions on the calculator. It should be clear that the largest enclosed area happens from C to D. Also, the cosine function is on top of the sin function from C to D. E

  40. #40 - Calculator Graph the function on the interval [0,1.4]. The zero on that interval is at x = 1.042. Then Math 8 at x = 1.042 to get 3.451. C

  41. #41 - Calculator The area will be approximated using the values at x = 1, 3, and 5. Setting up the Riemann sum gives (0.25)(2) + (0.68)(2) + (0.95)(2) = 3.76 D

  42. #42 – Calculator This is a direct application of the Fundamental Theorem of Calculus. Since we are taking the derivative of the integral we get the original function multiplied by the derivative of the interval of values. C

  43. #43 - Calculator The amount of water in the tank is given by After 4 hours this becomes D

  44. #44 - Calculator Solve 2 sin x = x to get x = 1.895. Then D

  45. #45 - Calculator Find the equation of the tangent line. . At x = 1, . The equation of the tangent line is . To find the area of the triangle you need to find the x-intercept and y-intercept to get the base and the height. The x-intercept is x = 2 and the y-intercept is y = 2e. The area is A

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