The diagram below shows the graphs of f(x) = 2sin 2x and g(x) = a tan bx. Give the coordinates of A and B. Write down the coordinates of C and D. What are the values of a and b?

Step 1: Determine the coordinates of A and B. Points A and B are the x-intercepts of the function f(x) = 2(2x). At x-intercepts, f(x) = 0. 2(2x) = 0 (2x) = 0 This occurs when 2x = n · 180^ for any integer n. x = n · 90^ From the graph, A is the first x-intercept to the left of the origin, so n=-1, which gives x = -90^. B is the first x-intercept to the right of the origin, so n=1, which gives x = 90^. The coordinates are: A: (-90^, 0) B: (90^, 0) Step 2: Determine the coordinates of C and D. Point C is a local maximum of f(x) = 2(2x). The maximum value of 2(2x) is 2. This occurs when (2x) = 1. 2x = 90^ + k · 360^ x = 45^ + k · 180^ From the graph, C is the first maximum in the positive x-axis, so k=0, which gives x = 45^. Point D is a local minimum of f(x) = 2(2x). The minimum value of 2(2x) is -2. This occurs when (2x) = -1. 2x = -90^ + k · 360^ x = -45^ + k · 180^ From the graph, D is the first minimum in the negative x-axis, so k=0, which gives x = -45^. The coordinates are: C: (45^, 2) D: (-45^, -2) Step 3: Determine the values of a and b. The function is g(x) = a(bx). From the graph, the vertical asymptotes of g(x) are at x = -90^ and x = 90^. For a tangent function, vertical asymptotes occur when the argument is 90^ + n · 180^. So, bx = 90^. Using x = 90^: b(90^) = 90^ b = 1 Now, use point C (45^, 2), which lies on g(x). Substitute x=45^, y=2, and b=1 into g(x): 2 = a(1 · 45^) 2 = a(45^) Since (45^) = 1: 2 = a(1) a = 2 The values are: a=2, b=1 Step 4: Determine the values of x for which f(x) > g(x) if x [0^; 90^]. We need to solve 2(2x) > 2(x) for x [0^; 90^]. Divide by 2: (2x) > (x) Use the identities (2x) = 2(x)(x) and (x) = ((x))/((x)): 2(x)(x) > ((x))/((x)) For x (0^, 90^), both (x) > 0 and (x) > 0. Multiply both sides by (x) (which is positive, so the inequality direction remains the same): 2(x)^2(x) > (x) Divide both sides by (x) (which is positive, so the inequality direction remains the same): 2^2(x) > 1 ^2(x) > (1)/(2) Since (x) > 0 in the interval (0^, 90^), take the positive square root: (x) > sqrt((1)/(2)) (x) > (1)/(sqrt(2)) (x) > sqrt(2)2 We know that (45^) = sqrt(2)2. Since (x) is a decreasing function in the interval (0^, 90^), for (x) to be greater than sqrt(2)2, x must be less than 45^. Also, at x=0^, f(0)=0 and g(0)=0, so f(0) g(0). At x=45^, f(45^)=2 and g(45^)=2, so f(45^) g(45^). Therefore, the inequality holds for x values strictly between 0^ and 45^. The values of x are: 0^ < x < 45^ Send me the next one 📸