AP Calculus AB Exam Section II, Part B Martha Torres, Tania Garcia, Michelle Samayoa

Problem #6: Let f be a function defined by f(x) ={1 – 2sinx for x ≤ 0 6A) Show that f is continuous at x = 0. 6B) For x ≠ 0, express f’(x) as a piecewise-defined function. Find the value of x for which f’(x) = -3. 6C) Find the average value of f on the interval [-1,1] e-4x for x ≤ 0

6A) Show that f is continuous at x = 0. Continuity: A function is said to be continuous on the interval [a, b] if it is continuous at each point in the interval. In order for f(x) to be continuous at x = 0, both equations of the function must meet at the same point at f(0). To prove that they have the same y value at x = 0, set both equations to f(0) and set them equal to each other.

e-4(x) = 1 – 2sinx e-4(0) = 1 – 2sin(0) e(0) = 1 – 2(0) 1 = 1 – 0 1=1 The function is continuous at x = 0 because at x = 0, the both have the same value: y = 1.

6B) For x≠ 0, express f’(x) as a piecewise-defined function. Find the value of x for which f’(x) = -3. f(x) ={1 – 2sinx for x ≤ 0 f’(x) ={-2cosx for x ≤ 0 e-4x for x ≤ 0 -4e-4x for x ≤ 0

_____ -4 6B) Continued. -2cosx = -3 cosx ≠ 3/2 -4e-4x = -3 e-4x = -3/-4 ln(e-4x)= ln(3/4) -4x = ln(3/4) x= ln(3/4)/-4 x= ln(3/4)

6C) Find the average value of f on the interval [-1,1] The average value of a function  over the interval [a,b] is given by:

6C) Continued (1/1-(-1)) ∫  f(x) dx = (1/2)∫  (1 – 2sinx) dx + ∫  (e-4x) dx = (1/2)[x + 2cosx] – [e-4x/4] = (1/2)[(0 + 2) – (-1 + 2cos(-1)) – ((e-4/4) – (1/4))] = (1/2)[(13/4) – (e-4/4) – 2cos(-1)] 1 -1 -0 -1 1 0 -0 -1 1 0