Math Key to Free Responses

SCORING GUIDELINES Question 1 Grass clippings are placed in a bin, where they decompose. For 0 � t � 30, the amount ofgrass clippings remaining in the bin is modeled by A(t) = 6.687( 0.931)', where A(t) is measured in pounds and t is measured in days. (a) Find the average rate ofchange ofA(t) over the interval O � t � 30. Indicate units ofmeasure. (b) Find the value ofA'(15). Using correct units, interpret the meaning ofthe value in the context ofthe problem. �ind the time t for which the amount ofgrass clippings in the bin is equal to the average amount ofgrass �ippings in the bin over the interval O � t � 30. (d) For t > 30, L(t), the linear approximation to A at t = 30, is a better model for the amount ofgrass clippings remaining in the bin. Use L(t) to predict the time at which there will be 0.5 pound ofgrass clippings remaining in the bin. Show the work that leads to your answer. (a) A(30)-A(0) = -0.197 (or -0.196) lbs/day 30 _ 0 (b ) A' (l5) = -0.164 (or -0.163) The amount ofgrass clippings in the bin is decreasing at a rate of0.164(or 0.163)lbs/day at time t = 15 days. 1 � 30 (d) L(t) J:0 A(t) dt ⇒ t = 12.415 (or 12.414) = A(30) + A'(30) • (t - 30) A'(30) = -0. 055976 A(30) = 0.782928 L(t) = 0.5 ⇒ t = 35. 054 1: answer with units . 1 : A'(15) 2•{ 1: interpretation I f3o I: J A(t)dt 2: { 30 o 1 : answer 2: expression for L(t) 4: { 1: L(t) = 0.5 I: answer SCORING GUIDELINES Question 5 No u.f �l�+o�. 1 , where k is a nonzero constant. The derivative off is given by Consider the function f(x)= 2 x -kx k - 2x . f'(x) = 2 (x2 - kx) 1 . Write an equation for the line tangent to the graph off at the point (a) Let k = 3, so that f(x)= 2 X -3x whose x-coordinate is 4. 1 . Determine whether f has a relative minimum, a relative maximum, or (b) Let k = 4, so that /( x) = 2 X -4X neither at x=2. Justify your answer. (c) Find the value of k for which f has a critical point at x = -5. 1 k =6, so that f( x)= 2 . Find the partial fraction decomposition for the function f X -6x f(x) dx. f An equation for the line tangent to the graph off at the point whose x-coordinate is 4 is y=- �( x - 4) + 1 4-2x (b) f'(x)= 2 (x2 -4x) -1--- /'(2)= ¼. 4-2·2 =O 2 (22 -4·2} -+ '2. �· f'(x) changes sign from positive to negative at x=2. Therefore, f has a relative maximum at x = 2. (c) f'(-5)= , k 2 ( ) - · -S =O⇒k=-10 ((-5) 2 - k • (-5) )2 I = = � + __!!._ ⇒ l = A(x-6) + Bx I -6x X( X-6) X X- 6 x= 0⇒ l = A •(-6 ) ⇒ A =--l 6 x=6⇒ 1=B·(6)⇒B=¼ = -J/ 6 + .Jlj__ I x-6 x x(x -6) - 6 6 f(x) dx = f ( 1! + �� ) dx 6 f = -¼lnlxl + ¼injx - 6 1 + C = ¼ml x:61 + C l : slope 2: { I : tangent line equation 2:{ l : considers /'(2) l : answer with justification I : answer 2 : partial fraction decomposition 4: { 2 : general antiderivative SCORING GUIDELINES Question 4 No e4\w.lA-"0f. y The figure above shows the graph of/', the derivative of a twice-differentiable function f, on the closed interval 0 � x ::;; 8. The graph of f' has horizontal tangent lines at x = I, x = 3, and x = 5. The areas of the regions between the graph of J' and the x-axis are labeled in the figure. The function f is defined for all real numbers and satisfies /(8} = 4. (8, 5) (a) Find all values of x on the open interval O < x < 8 for which the function f has a local minimum. (5.-2) Justify your answer. Graph of J' �etennine the absolute minimum value ofJon the )"closed interval O � x � 8. Justify your answer. (c) On what open intervals contained in O < x < 8 is the graph off both concave down and increasing? Explain your reasoning. (d) The function g is defined by g(x) = (f(x))3 . If /(3} = -{, find the slope of the line tangent to the graph of g at x = 3. (a) x = 6 is the only critical point at which f' changes sign from negative to positive. Therefore, f has a local minimum at X = 6. �om part (a), the absolute minimum occurs either at �n endpoint. x = 6 or /(0) = /(8) + J ° J'(x) dx 8 I : answer with justification 1 : considers x = 0 and x = 6 3 : { I : answer I : justification = (8)- rso J, (x) dx = 4 -12 = -8 J / /(6) = /(8) + J: f'(x) dx = /(8) - J: 1·cx) dx = 4 - 1 = -3 /(8) = 4 The absolute minimum value off on the closed interval [0, 8] is -8. (c) The graph off is concave down and increasing on O < x < l and 3 < x < 4, because f' is decreasing and positive on these intervals. I : answer 2: { I : explanation (d) g'(x) = 3[J(x)]2 • f'(x) . 2: g'( x) 3 { • l : answer g'( 3) = 3[ (3) 2 (3) / ] • /' = 3(-% r. 4 = 75 SCORING GUIDELINES Question 6 No u-1""-la+of'. oo (-3)n-l (-3)n-l The Maclaurin series for a function f is given by L--'----'---Xn = x -1 x2 + 3x3 - • • • + --'---'---Xn + • • • and n n 2 n=I converges to f(x) for I xl < R, where R is the radius of convergence of the Maclaurin series. �e the ratio test to find R. (b) Write the first four nonzero terms of the Maclaurin series for f', the derivative off Express f' as a rational function for lxl < R. (c) Write the first four nonzero terms of the Maclaurin series for ex . Use the Maclaurin series for ex to write the third-degree Taylor polynomial for g(x) = exf(x) about x = 0. I c,k� ! lr �� an be the nth term of the Maclaurin series. /� ( 3)n x11+1 an+I n -3n x --=��-------=--· n n +1 n + 1 a,, ( 3r-1 x 1- 1 : sets up ratio 3 : { 1 : computes limit of ratio 1 : determines radius of convergence 3 •x n l = 3lxl lim n ➔oo n + 1 The radius of convergence is R = ½- (b) The first four non.zero terms of the Maclaurin series for f' are ) - 3x + 9x - 27x : ·:�-'\ 2 • , • 3 3:{ 1 .. •• 2 : first four nonzero terms 1 : rational function (x) = = IA'J' 1 -(-3x) I+ 3x (c) The first four nonzero terms of the Maclaurin series for ex are x2 + x3 ✓ 1 + X + 2! 3T· The product of the Maclaurin series for ex and the Maclaurin series for f is x (1 + x + x� + : + 2. 3. •··)(x _ ix 2 2 = x - _!_ x2 + 2x3 + · · · 2 + 3x3 - •··) , The third-degree Taylor polynomial for g(x) = exf(x) about x = 0 is T3 (x) = x -½x2 + 2x3 . L I 3:{ 1 : first four nonzero terms of the Maclaurin series for ex 2 : Taylor polynomial