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Calculus Lesson 2.4 Solutions 2-4, 8-12, 15-16 In Exercises 2-4, find the average rate of change of the function over each interval. 2) f(x) = 4x + 1 (a) [0, 2] (b) [10, 12] 3) f(x) = ex (a) [-2, 0] (b) [1, 3] 4) f(x) = ln x (a) [1, 4] (b) [100, 103] In Exercise 8, a distance-time graph is shown. (a) Estimate the slopes of the secants PQ1, PQ2, PQ3, and PQ4, arranging them in order in a table. What is the appropriate unit for these slopes? (b) Estimate the speed at point P. 8) The accompanying figure shows a distance-time graph for a wrench that fell from the top platform of a communication mast on the moon to the station roof 80 meters below. P 80 Q4 Q3 60 Q2 40 20 Q1 5 10 1 In Exercises 9-12, at the indicated point, find (a) the slope of the curve (b) an equation of the tangent, and (c) an equation of the normal line (d) Then draw a graph of the curve, tangent line, and normal line in the same square viewing window. 9) y = x2 at x = -2 10) y = x2 - 4x, at x = 1 11) y= 12) y = x2 - 3x - 1, at x = 0 1 , at x = 2 x-1 In Exercises 15-16, determine whether the curve has a tangent at the indicated point. If it does, give its slope. If not, explain why not. 2 15) f(x) = 2 - 2x - x , 2x + 2, 16) f(x) = -x, x2 - x, x<0 x≥0 x<0 x≥0 at x = 0 at x = 0 2 Answer Key Testname: L. 2.4 SOLUTIONS 2) 9- 1 =1 2 (a) f(2) - f(0) = 2-0 (b) f(12) - f(10) = 12 - 10 49 - 41 7 - 41 ‘ 0.298 = 2 2 3) (a) f(0) - f(-2) e0 - e(-2) 1 - e(-2) ‘ 0.432 = = 2 2 0 - (-2) (b) f(3) - f(1) e3 - e(1) ‘ 8.684 = 3-1 2 4) (a) f(4) - f(1) ln 4 - ln 1 ln 4 ‘ 0.462 = = 3 3 4-1 (b) f(103) - f(100) ln 103 - ln 100 ‘ 0.010 = 103 - 100 3 8) 80 - 20 ‘ 12.000 (a) PQ1 = 10 - 5 80 - 39 ‘ 13.667 PQ2 = 10 - 7 80 - 57 ‘ 15.333 PQ3 = 10 - 8.5 80 - 71 ‘ 18.000 PQ4 = 10 - 9.5 What is the appropriate unit for these slopes? meters per second (b) Estimate the speed at point P. Approximately 18 meters per second 3 Answer Key Testname: L. 2.4 SOLUTIONS 9) a) m = lim h¬0 f(-2 + h) - f(-2) h m = lim h¬0 (-2 + h)2 - (-2)2 h m = lim h¬0 4 + -4h + h 2 - 4 h m = lim h¬0 -4h + h 2 = lim -4 + h = -4 h h¬0 b) when x = -2, y = 4 y = mx + b 4 = (-4)(-2) + b 4=8+b b = -4 y = -4x - 4 c) m = 1 4 y-4= 1 (x + 2) 4 y-4= 1 1 x+ 4 2 y= 1 1 x+4 4 2 d) 7 y 6 5 4 3 2 1 -7 -6 -5 -4 -3 -2 -1 -1 1 2 3 x -2 -3 4 Answer Key Testname: L. 2.4 SOLUTIONS 10) a) m = lim h¬0 f(1 + h) - f(1) h m = lim h¬0 (1 + h)2 - 4(1 + h) - ((1)2 - 4(1)) h m = lim h¬0 1 + 2h + h2 - 4 - 4h - 1 + 4 h m = lim h¬0 2h + h2 - 4h = lim 2 + h - 4 = -2 h h¬0 b) when x = 1, y = -3 y = mx + b -3 = (-2)(1) + b b = -1 y = -2x - 1 c) m = 1 2 y+3= 1 (x - 1) 2 y+3= 1 1 x2 2 y= 1 1 x-3 2 2 d) 6 y 5 4 3 2 1 -5 -4 -3 -2 -1 -1 1 2 3 4 5 x -2 -3 -4 -5 -6 5 Answer Key Testname: L. 2.4 SOLUTIONS 6 y 6 Answer Key Testname: L. 2.4 SOLUTIONS 11) a) m = lim h¬0 m = lim h¬ 0 f(2 + h) - f(2) h 1 1 2+h-1 2-1 h 1 -1 1+h m = lim h¬0 h 1 1+h 1 1 m = lim = lim h¬0 (1 + h)h h h¬0 (1 + h)h (1 + h)h -h -1 = lim = lim = -1 h¬0 (1 + h)h h¬0 (1 + h) b) when x = 2, y = 1 y = mx + b 1 = (-1)(2) + b b=3 y = -1x + 3 c) m = 1 y - 1 = 1(x - 2) y - 1 = 1x - 2 y=x-1 d) 6 y 5 4 3 2 1 -5 -4 -3 -2 -1 -1 1 2 3 4 5 x -2 -3 -4 -5 -6 7 Answer Key Testname: L. 2.4 SOLUTIONS 6 y 8 Answer Key Testname: L. 2.4 SOLUTIONS 12) a) m = lim h¬0 f(0 + h) - f(0) h m = lim h¬0 (h2 - 3(h) - 1) - (02 - 3(0) - 1) h m = lim h¬0 h2 - 3h - 1 + 1 h m = lim h¬0 h2 - 3h = lim h - 3 = -3 h h¬0 b) when x = 0, y = -1 y = mx + b -1 = (-3)(0) + b b = -1 y = -3x - 1 c) m = 1 3 y+1= 1 (x - 0) 3 y+1= 1 x 3 y= 1 x-1 3 d) 6 y 5 4 3 2 1 -5 -4 -3 -2 -1 -1 1 2 3 4 5 x -2 -3 -4 -5 -6 9 Answer Key Testname: L. 2.4 SOLUTIONS 6 y 5 (2 - 2(h + 0) - (h + 0)2) - (2 - 2(0) - (0)2) = lim h h¬0= lim h ¬ 0- = lim h¬0+ = lim h ¬ 0+ (2 - 2h - h2) - 2 = -2 - h = -2 h (2(h + 0) + 2) - (2(0) + 2) h (2h + 2) - 2) =2 h The tangent line does not exist. The slope from the left is different than the slope from the right. 6 y 5 4 3 2 1 -5 -4 -3 -2 -1 -1 1 2 3 4 5 x -2 -3 -4 -5 -6 10 Answer Key Testname: L. 2.4 SOLUTIONS 16) = lim h ¬ 0- = lim h ¬ 0+ = lim h¬0+ -(h + 0) - (-0) = -1 h ((h + 0)2 - (h + 0)) - ((0)2 - 0) h h2 - h = h 6 lim h - 1 = -1 h¬0+ y 5 4 3 2 1 -5 -4 -3 -2 -1 -1 1 2 3 4 5 x -2 -3 -4 -5 -6 11