The Function f(x) is Twice Differentiable, and the Graph of f (x) has no Points of Inflection

1. The function f x is twice differentiable, and the graph of f x has no points of inflection. If f 6 = 3, -1 , and f ¢ ¢ 6 = -2, which of the following could be the value of f 7 ? f ¢ 6 = 2 A B 2 C 2.5 D 2.9 E 3 4 If we use Euler ' s Method, we see that using the initial point 6, 3 and a step size of 1, the next -1 approximation would be 3 + 1 = 2.5 and the point would be 7, 2.5. Since f ¢ ¢ 6 = -2 2 and there are no points of inflection, this function is always concave down. This means that the value at x = 7 must be less than 2.5, so the only possibility is a value of 2, and the answer is A 2. Consider a function f x, where f x > 0 on a, b, f ¢ x > 0 on a, b, and f ¢ ¢ x < 0 on a, b If the interval a, b is split into 4 equal subdivisions, and L is the left - hand Riemann sum for the partitioning, R is the right - hand Riemann sum for the partitioning, and T is the trapezoidal approximation for the partitioning, then A B L < T < R C T < R < L D R < T < L E L < R < T R < L < T The described function is positive, increasing, and concave down over the given interval. If we were to draw the right - hand Riemann sum, we would see that these rectangles overapproximate the area. The trapezoids over the same intervals would be an underapproximation, while the left - hand Riemann sum would be an even worse underapproximation, so the correct ordering is L < T < R and the answer is A 2  3. Evaluate 1 A x + 1 x 1 1 2 2 + 2x 8  1 u 3 du dx x2 + 2 x B ln 8 - ln 3 2  x + 1 1 = 2 2 ln u  8 ln 8 - ln 3 = D ln 8 and du = 2 x + 2 dx u = x2 + 2 x Æ dx C ln 8 - ln 3 2 2 du = x + 1 dx and the answer is 2 3 1 or E 3 ln 2 3 ln 2 + 2 2 so we have B 4. The base of a solid is the area in the first quadrant bounded by the x - axis, the y - axis, and the circle x2 + y2 = 4. Every cross section of the solid is an equilateral triangle, perpendicular to the x - axis, with one side in the xy - plane. A B 16 p 3 2 V =  A x dx C 4 3 where A x = 0 A x = 3 4 Find the volume of the solid 4 - x 2 and so 3 2 = V = 4 3 4 4 D 16 3 side2 3 4 - x  so V =  3 4 0 8 - 8 3 = 4 3 3 4 16 3 = 4 - x2  dx 4 3 3 4 E 3 = 3 4 4 x - and the answer is s2 . 3 2 4 - x2 and a side is top - bottom, or side = 2 2 3 area of an equilateral triangle is 1 3 D so x3  2 0 5. If y = f x3  A 9 x2 - 2 D If dy dx 3 x2 f ¢ x = and 3x - 2 B 3 x3 - 2 y = f x3  = 3 x2 dy then dx 3 x3 - 2 x3 E = f ¢ x3  3 x2 6. Consider the graph below of f x, dx = C 3 x3 - 2 18 x3 and and the answer is dy then 3 x2 9 x2 - 2 9 x2 - 2 f ¢ x3  = 3 x3  - 2 so dy 3 x3  - 2 3 x2 = dx D defined on the interval 0, 4. x Let G x =  f t dt and 0 x H x =  f t dt. Which of the statements below is true ? 2 y 2 fx 1 1 2 3 4 x -1 A D -2 G x = H x G x = H x - 2 B G ¢ x = H ¢ x + 2 E G x = H x + 2 C x Notice that the area under the curve from 0 to 2 is or G x = H x + 3 3. Or, G x =  f t dt 0 and the answer is C G x = H x + 3 2 = x  f t dt +  f t dt 0 2 or 7. Suppose that the derivative of f has the graph as shown below y f  x x 1 Which of the following could be the graph of f ? y y A y B x C x y x y D E x By inspection, the answer is x C b 8. If f is a linear function and 0 < a < b, then ¢¢  f x dx = a A B 0 b ¢¢  f x dx = f ¢ x  a = b a m - m C 1 D ab 2 but since f x = mx + b, = 0 then f ¢ x = m and the answer is E b - a so f ¢ x  b = a b2 - a2 2 m  b a A 9. Suppose that f x, f ¢ x, and f ¢¢ x are continuous for all real numbers x, and that following properties i f is positive on -•, 4 and negative on 4, • ii f is increasing on -•, -1 and decreasing on -1, • iii f is concave down on -•, 2 and concave up on 2, • f has the Of the following, which has the greatest numerical value? A f 5 B f ¢ ¢ 1 C f ¢ 0 All of the given values are negative except D f ¢ -2 f ¢ -2 E Not enough information is given and the answer is D 10. If the base b of a triangle is increasing at a rate of 3 inches per minute while its height h is decreasing at a rate of 3 inches per minute, which of the following must be true about the area A of the triangle ? A D 1 A = 2 bh dA so = dt and the answer is 1 db 2 dt h + b dh A is always decreasing C A remains constant E A is decreasing only when b < h 1 = dt 2 3 h - 3 b < 0 Æ 3h < 3b or when b > h D 11. On the interval A D B A is always increasing A is decreasing only when b > h 0, p 2 , y = sin xx , if x ln sin x sin xx x cos x + sin x B then dx = sin xx cot x C x sin xx - 1 cos x x E sin x x cot x + ln sin x ln y = x ln sin x, Using logarithmic differentiation, 1 ¢ 1 y = ln sin x + x cos x y sin x dy and we differentiate both sides to get y ¢ = sin xx ln sin x + x cot x or and the answer is 12. If the curve below represents the complete polynomial graph of f x, then the graph of f ¢ ¢ x the x - axis in how many points ? 4 E will cross y 3 2 fx 1 2 A 2 4 6 B 3 C 4 Count the points of inflection. A x sin-1 x + D x sin-1 x - 2 D 5 E 6 There are 4 points of inflection B 1 - x2 + C u = sin-1 x x sin-1 x E 1 - x2 + C 1 du = = x sin-1 x -  x 1 - x x = -17 12 C B x = -9 8 C 1 - x2 + C 1 2 v = x x sin-1 x + 1 1 - x2 + C 2 1 - x2 + C dv = dx so 2 dx = x sin-1 x + 1 - x2 + C and the answer is A 2 14. Which of the following is a critical value for the function or zero, from the first derivative. A and the answer is x sin-1 x - dx 1 - x -1  sin x dx x -1  sin x dx 13. Evaluate -1  sin x dx 8 C f x = 2 x + 33 4 x + 33 ? x = -1 D x = -9 10 This would be a root, E x = 17 12 f ¢ x = 3 2 x + 32 2 4 x + 33 + 2 x + 33 3 4 x + 32 4 so 2 2 x + 32 4 x + 32 48 x + 54 -3 -3 -9 so the critical values are x = , , and 2 4 8 2 f ¢ x = 2 x + 32 4 x + 32 6 8 x + 9 and the answer is 15. If f x = 3 x ¢ A If 2 B 3  x + 9 dx 0  = 0 1 4 0 C f x = x3 + C then  f x dx then find 22 2 = 16 1 = -23 + C and x4 + 9 x  2 D = 0 E 9 C = 9 so 4 + 18 - 0 = 22 An equation of the normal line to the graph of f x = x3 + 3 x2 + 7 x - 1 x = -1 is 16. A D B x + 4 y = 25 E 4 x + y = -10 x + 4 y = -25 x - 4 y = 23 f ¢ x = 3 x2 + 6 x + 7 f -1 = -1 + 3 - 7 - 1 or y + 6 = Dx x + 1 -1 4 f ¢ -1 = 3 - 6 + 7 and and 17. Find 1 g x = -6 Æ x= 5 4 5 if 4 g x and and the answer is C 4x - y = 2 x + 4 y = -25 y - -6 = is defined by the following graph gx 3 2 1 x A Dx 3 B -1 2 1 g x - g x = 8 D 1 8 and at x = 5, we have - g 5 g 52 E 1 4 ¢ g x2 18. If a = b e2 x , ln 7 C ¢ and the answer is A 6 -1 = --1  2 22 = 1 2 1 8 C and ln a = 7 B 2 and ln b = 3, C 7 3 then find x D 4 -1 4 and the answer is y 2 m = so the slope of the normal line is Æ B at the point where so the equation of the normal line is -4 y - 24 = x + 1 at = 4 4 1 or 2 f -2 = 1, and f ¢ x = 3 x2  f x dx = B 28 2 2 x + 32 4 x + 32 6 4 x + 3 + 12 2 x + 3 = f x = 2 x + 3 4 x + 3 24 x + 18 + 24 x + 36 ¢ E e3 -1 4 x - -1 A ln a = ln b e2 x  Take the natural log of both sides to get 7 = 3 + 2x we have so x = 2 ln a = ln b + 2 x or and the answer is and substituting B 19. The slope field below matches which differential equation ? 3 2 1 -3 -2 -1 1 2 3 -1 -2 -3 A dy dx D dy dx = = B x 2y E x y + 1 dy dx dy dx 2 = = C 2y x dy dx 1 = 3 x2 + 1 1 xy Notice that, on the y - axis, the slopes are zero. This suggests that x should be a factor, and narrows the choices to either A or D. Testing a few other order pairs on the x - axis, we see that the correct choice is D 20. Suppose that f ¢ ¢ x > 0 A B C D E throughout the domain of a given function. If we are given dy dx , an initial value, and asked to use Euler¢ s Method to approximate a solution, then Euler ¢ s Method will underapproximate the solution Euler ¢ s Method will overapproximate the solution Euler ¢ s Method will underapproximate the solution for all x > 0, overapproximate for all x £ 0 Euler ¢ s Method will overapproximate the solution for all x > 0, underapproximate for all x £ 0 Not enough information is given to make this determination Since the second derivative is everywhere positive, the tangent line used will be under the curve, and Euler ' s Method will underapproximate the solution, and the answer is A 21. Let f be the function defined by value of g ¢ 2. A B 1 13 Since f 1 = 2, f x = x3 + x. 1 4 and f ¢ x = 3 x2 + 1, C 0 B 4 9 D 7 4 then f ¢ 1 = 4, 22. Find the slope of the line tangent to the curve A If g x = f -1 x C 7 9 so g ¢ 2 = 6 7 g 2 = 1, E 4 3 y2 - 2 x2 = 6 - 2 x y D and 1 4 then find the 13 and the answer is at the point 3, 2. E 5 3 B Differentiating implicitly, we have 6 y y ¢ - 4 x = -2 y + x y ¢ 12 y ¢ - 12 = -2 2 + 3 y ¢  and the answer is A D x2 + 2 x + 1 = Æ 18 y ¢ = 8 dy dx x = 1, x2 + 2 x + 1 but the function and the answer is y = 8 ex x + 1 B = x + 12 9 has a removable discontinuity x + 1 y = ln x + 1 + 8 = 3 x2 y 1 dy Æ E = 3 x2 y dx C = ln 8 A B -3 = k + x2 k2 - 8 k + -2 2 = 1 = k2 - 8 Æ y dx y = C -1 k + x k - 1 k x + 8 1 dy = dy ln y = x3 + ln 8 or 3 y = ex + 7 y2 = x3 + 8  Æ If the curve contains the point C y = x3 + 8 25. The equation of the line tangent to the curve dx 18 D 3 and ln 8 = 0 + C so and the answer is A dy 4 = has a jump discontinuity C has an infinite discontinuity E has both a removable and infinite discontinuity 24. At each point x, y on a certain curve, the slope of the curve is 3 x2 y. 0, 8, then the equation of the curve is D 8 y¢ = x< 1 x + 1 is continuous B has a removable discontinuity x = -1, A Æ 2x x ≥ 1 The best description concerning the continuity of g x is that the function g x is continuous at at 12 y ¢ - 12 = -4 - 6 y ¢ Æ B g x 23. Suppose and plugging in, we get kx + 8 k + x 1 k2 + k x - kx - 8 k + x2 = k2 - 4 k + 4 = Æ 2  3 x dx y = ex Æ at Æ 3 x = -2 D ln y + ln 8 is or E k2 - 8 k + x2 and at or y = 8 ex y = x + 4. 3 4 k = 12 = x3 + C k = 3 3 The value of k is 4 x = -2, dy dx = 1 so and the answer is 26. For values of h very close to 0, which of the following functions best approximates csc x + h - csc x f x = ? h A B csc x We are looking for so 27. A d dx lim xÆ3 -6 x2 - 9 x - 3 h csc x + h - csc x h hÆ0 csc x = -csc x cot x lim C csc x - csc x D E -csc x h - csc x cot x and this is the definition of the derivative of csc x, and the answer is E = B -1 C 1 D 6 E Does Not Exist D lim xÆ3 x2 - 9 = x - 3 + x2 - 9 lim - xÆ3 = x - 3 xÆ3 lim x - 3 x + 3 x - 3 x + 3 lim - Because these two limits do not match, 28. Find Dx ln h x2    A B 1 4 Dx ln h x2    = or 6 4 2 = 1 h x  2 if h 3 = 2, x = 3 at C 1 2 h ¢ x2  2 x 3 and at and the answer is 29. Given the graphs of lim -1 x + 3 h 9 = 4, D 2 1 h 3  Æ 2 h ¢ 32  2 3 3 1 h 9 = h ¢ 9 6 y hx x x f ¢ x = g x and f ¢ ¢ x = h x h ¢ x = g x and h ¢ ¢ x = f x g ¢ x = h x and g ¢ ¢ x = f x 30. If f x = cos x - sin x, f x has a local maximum at x = C f x is increasing on  E f x has a local minimum at 4 f ¢ x = h x h ¢ x = f x f ¢ ¢ x = g x h ¢ ¢ x = g x and and and note that f ¢ x = g x, and h ¢ x = f x then which of the following is true on the interval A 3p x B D Focus on the slopes of the functions at x = 0, and the answer is D , f ¢ x = -sin x - cos x = 0 7p 4  B 3p 4 D only x = Æ and f ¢ x is positive on and 3p f x is decreasing on  4 3p 4 p x = 5p , 4 4  only 5p 4 -sin x = cos x , 0, 2 p ? f x has a local minimum at Æ -tan x = 1 and setting up a sign line, we see that f ¢ x is negative on 0, 4 E E gx 3p E and h ¢ 9 = 2 2 y fx LMins at x = and the answer is h ¢ 3 = 1, 3 x = 3 -6 f x, g x, and h x, which coul be true ? y A C E = Does Not Exist x - 3 xÆ3 6 x Æ 3+ x2 - 9 lim = x Æ 3+ = x - 3 xÆ3 lim x + 3 = x - 3 + 7p 4 x = 2p 3p 4 Æ and with LMaxes at x = 0 and x = and the answer is C c = 7p 4 7p 4 3p 4 , 2 p and , 7p 4