In the diagram, f(x) = 2 (x-45^) is drawn for the interval x [-180^ ; 180^]. Write down the amplitude of f.

QUESTION 7 7.1 Write down the amplitude of f. The function is given by f(x) = 2 (x - 45^). For a sine function of the form y = A (Bx + C) + D, the amplitude is |A|. In this case, A = 2. The amplitude of f is 2. 2 7.2 Determine the values of x, in the interval x [-180^; 180^], for which f'(x) 0. Step 1: Find the derivative f'(x). Given f(x) = 2 (x - 45^). The derivative is f'(x) = (d)/(dx) [2 (x - 45^)] = 2 (x - 45^). Step 2: Set f'(x) 0. We need to solve 2 (x - 45^) 0, which simplifies to (x - 45^) 0. Step 3: Find the general solution for 0. Let = x - 45^. The cosine function is non-negative in the first and fourth quadrants. So, -90^ + 360^ n 90^ + 360^ n, where n is an integer. Step 4: Substitute back = x - 45^ and solve for x. -90^ + 360^ n x - 45^ 90^ + 360^ n Add 45^ to all parts: -90^ + 45^ + 360^ n x 90^ + 45^ + 360^ n -45^ + 360^ n x 135^ + 360^ n Step 5: Find the values of x in the interval x [-180^; 180^]. For n=0: -45^ x 135^ This interval is within [-180^; 180^]. For other integer values of n, the intervals will fall outside [-180^; 180^]. -45^ x 135^ 7.3 On the same set of axes provided in the ANSWER BOOK, draw the graph of g(x) = - x for the interval x [-180^; 180^]. To draw g(x) = - x: • Asymptotes: The asymptotes for x are at x = 90^ + 180^ n. For g(x) = - x, the asymptotes are the same. In the interval [-180^; 180^], the asymptotes are at x = -90^ and x = 90^. Draw vertical dashed lines at these x-values. • Key points: • g(-180^) = -(-180^) = 0 • g(-135^) = -(-135^) = -(-1) = 1 • g(-45^) = -(-45^) = -(-1) = 1 • g(0^) = -(0^) = 0 • g(45^) = -(45^) = -1 • g(135^) = -(135^) = -(-1) = 1 • g(180^) = -(180^) = 0 • Shape: The graph of g(x) = - x is a reflection of x about the x-axis. It will decrease from positive infinity to negative infinity between asymptotes. • From x = -180^ to x = -90^: The graph starts at ( -180^, 0) and decreases towards - as it approaches x = -90^. • From x = -90^ to x = 90^: The graph comes from as it leaves x = -90^, passes through (-45^, 1), (0^, 0), (45^, -1), and decreases towards - as it approaches x = 90^. • From x = 90^ to x = 180^: The graph comes from as it leaves x = 90^, passes through (135^, 1), and decreases to (180^, 0). 7.4 Write down the value of g(-45^) - f(-45^). Step 1: Calculate g(-45^). g(x) = - x g(-45^) = -(-45^) = -(-1) = 1 Step 2: Calculate f(-45^). f(x) = 2 (x - 45^) f(-45^) = 2 (-45^ - 45^) = 2 (-90^) = 2(-1) = -2 Step 3: Calculate the difference. g(-45^) - f(-45^) = 1 - (-2) = 1 + 2 = 3 3 7.5 How many solutions does the equation g(x) - f(x) = 0 have in the interval x [-180^; 180^]? The equation g(x) - f(x) = 0 is equivalent to g(x) = f(x). We need to find the number of intersection points between the graphs of f(x) and g(x) in the given interval. Let's analyze the graphs: • f(x) = 2 (x - 45^): • f(-180^) = 2 (-225^) = 2(sqrt(2)2) = sqrt(2) ≈ 1.414 • f(-135^) = 2 (-180^) = 0 • f(-45^) = 2 (-90^) = -2 • f(0^) = 2 (-45^) = -sqrt(2) ≈ -1.414 • f(45^) = 2 (0^) = 0 • f(135^) = 2 (90^) = 2 • f(180^) = 2 (135^) = sqrt(2) ≈ 1.414 • g(x) = - x: • g(-180^) = 0 • g(-135^) = 1 • g(-90^) is undefined (asymptote) • g(-45^) = 1 • g(0^) = 0 • g(45^) = -1 • g(90^) is undefined (asymptote) • g(135^) = 1 • g(180^) = 0 Let's examine the intervals: 1. x [-180^, -90^): • At x=-180^, f(-180^) = sqrt(2) and g(-180^) = 0. So f > g. • As x -90^, f(x) -sqrt(2) and g(x) -. So f > g. • In this interval, f(x) decreases from sqrt(2) to -sqrt(2), while g(x) decreases from 0 to -. f(x) remains above g(x). No intersections. 2. x (-90^, 90^): • As x -90^, f(x) -sqrt(2) and g(x) . So g > f. • At x=0^, f(0^) = -sqrt(2) and g(0^) = 0. So g > f. • At x=45^, f(45^) = 0 and g(45^) = -1. So f > g. • Since g > f at 0^ and f > g at 45^, there must be one intersection between 0^ and 45^. • As x 90^, f(x) sqrt(2) and g(x) -. So f > g. • In (-90^, 0^), g(x) is positive and f(x) is negative. No intersections. • In (45^, 90^), g(x) is negative and f(x) is positive. No intersections. 3. x (90^, 180^]: • As x 90^, f(x) sqrt(2) and g(x) . So g > f. • At x=135^, f(135^) = 2 and g(135^) = 1. So f > g. • Since g > f as x 90^ and f > g at 135^, there must be one intersection between 90^ and 135^. • At x=180^, f(180^) = sqrt(2) and g(180^) = 0. So f > g. • In (135^, 180^], f(x) decreases from 2 to sqrt(2), while g(x) decreases from 1 to 0. Since f(x) is always above g(x) at 135^ and 180^, and both are decreasing, f(x) remains above g(x). No intersections. Total number of solutions = 0 + 1 + 1 = 2. 2 solutions 7.6 Graph h is obtained when g is translated 45^ to the right. Write down the equations of the asymptotes of h in the interval x [-180^; 180^]. Step 1: Determine the equation of h(x). The graph g(x) = - x is translated 45^ to the right. This means replacing x with (x - 45^). So, h(x) = -(x - 45^). Step 2: Find the asymptotes of h(x). The asymptotes of are at = 90^ + 180^ n. For h(x), we set x - 45^ = 90^ + 180^ n. x = 90^ + 45^ + 180^ n x = 135^ + 180^ n Step 3: Find the asymptotes in the interval x [-180^; 180^]. • For n=0: x = 135^. This is in the interval. • For n=-1: x = 135^ - 180^ = -45^. This is in the interval. • For n=1: x = 135^ + 180^ = 315^. This is outside the interval. • For n=-2: x = 135^ - 360^ = -225^. This is outside the interval. The equations of the asymptotes of h in the given interval are x = -45^ and x = 135^. x = -45^ and x = 135^ What's next?