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Problem 1: First derivative: Productrule Problem 2: First derivative and Domain of Definition Problem 3: First derivative Problem 4: First derivative Problem 5: Local Maxima/Minima T(x) = -x² + 4x + 4 T(x) = 1 2 x x4 4 T(x) = -2,25x² + 4,5x +18 T(x) = 1 2 x 2x 3 2 T(x) = 2x²-6x + 65 -(x - 2)² + 8 TMax = 8 für x = 2 1 ( x 2)² 5 4 TMax = 5 für x = 2 -2,25 (x -1)² + 20,25 TMax = 20,25 für x = 1 1 ( x 2)² 1 2 TMin = 1 für x = 2 2(x-1,5)² + 60,5 TMin = 60,5 für x = 1,5 T(x) = 2 2 x 3 5 2 ( x 0) 2 3 5 TMin = -3 für x = 0 T(x) = 1 2 x 2x 1 2 1 ( x 2)² 1 2 TMin = -1 für x = -2 T(x) = 1,5 x2 - 5 1,5 (x - 0)2 - 5 TMin = -5 für x = 0 T(x) = -0,25x² + 3x -5,25 -0,25(x - 6)² + 3,75 TMax = 3,75 für x = 6 4 ( x 2,5)² 120,33 3 TMax =120,33 für x= 2,5 -(x -3)² + 7 TMax = 7 für x = 3 T(x) = 4 2 20 x x 112 3 3 T(x) = -x² + 6x - 2 Problem 6: Tangent Line Problem 1: Find all points on the graph of y = x 3 - 3x where the tangent line is parallel to the x axis (or horizontal tangent line). Solution to Problem 1: Lines that are parallel to the x axis have slope = 0. The slope of a tangent line to the graph of y = x 3 - 3x is given by the first derivative y '. y ' = 3x 2 - 3 We now find all values of x for which y ' = 0. 3x 2 - 3 = 0 Solve the above equation for x to obtain the solutions. x = -1 and x = 1 The above values of x are the x coordinates of the points where the tangent lines are parallel to the x axis. Find the y coordinates of these points using y = x 3 - 3x for x = -1 , y = 2 for x = 1 , y = -2 The points at which the tangent lines are parallel to the x axis are: (-1,2) and (1,-2). See the graph of y = x3 - 3x below with the tangent lines. Problem 2: Find a and b so that the line y = -3x + 4 is tangent to the graph of y = ax3 + bx at x = 1. Solution to Problem 2: We need to determine two algebraic equations in order to find a and b. Since the point of tangency is on the graph of y = ax3 + bx and y = -3x + 4, at x = 1 we have a(1)3 + b(1) = -3(1) + 4 Simplify to write an equation in a and b a+b=1 The slope of the tangent line is -3 which is also equal to the first derivative y ' of y = ax3 + bx at x = 1 y ' = 3ax2 + x = -3 at x = 1. The above gives a second equation in a and b 3a + b = -3 Solve the system of equations a + b = 1 and 3a + b = -3 to find a and b a = -2 and b = 3. See graphs of y = ax3 + bx, with a = -2 and b = 3, and y = -3x + 4 below. Problem 3: Find conditions on a and b so that the graph of y = ae x + bx has NO tangent line parallel to the x axis (horizontal tangent). Solution to Problem 3: The slope of a tangent line is given by the first derivative y ' of y = ae x + bx. Find y ' y ' = ae x + b To find the x coordinate of a point at which the tangent line to the graph of y is horizontal, solve y ' = 0 for x (slope of a horizontal line = 0) ae x + b = 0 Rewrite the above equation as follows e x = -b/a The above equation has solutions for -a/b >0. Hence, the graph of y = ae x + bx has NO horizontal tangent line if -a/b <= 0 Problem 7: Tangent Line 1 - Find all points on the graph of y = x 3 - 3x where the tangent line is parallel to the line whose equation is given by y = 9x + 4. 2 - Find a and b so that the line y = -2 is tangent to the graph of y = ax2 + bx at x = 1. 3 - Find conditions on a, b and c so that the graph of y = ax 3 + bx 2 + cx has ONE tangent line parallel to the x axis (horizontal tangent). solutions to the above exercises 1- (2,2) and (-2,-2) 2- a = 2 and b = - 4 3- 4b 2 - 12 ac = 0 Problem 8: Limits (Grenzwerte) Find the limits of f(x), where x-> -∞ and x-> ∞ Problem 9: Limits (Grenzwerte)