The sketch below shows the graph of g(x) = x³ - 6x - 1. P and Q are the turning points and R the y-intercept of g.

The given function is g(x) = x^3 - 6x - 1. 9.1 Determine the coordinates of R. R is the y-intercept, which occurs when x=0. Step 1: Substitute x=0 into the function g(x). g(0) = (0)^3 - 6(0) - 1 g(0) = -1 The coordinates of R are (0, -1). 9.2 Determine the coordinates of the turning points P and Q. Turning points occur where the first derivative g'(x) is equal to zero. Step 1: Find the first derivative of g(x). g'(x) = (d)/(dx)(x^3 - 6x - 1) = 3x^2 - 6 Step 2: Set g'(x) = 0 and solve for x. 3x^2 - 6 = 0 3x^2 = 6 x^2 = 2 x = ± sqrt(2) Step 3: Substitute these x values back into g(x) to find the corresponding y values. For x = sqrt(2): g(sqrt(2)) = (sqrt(2))^3 - 6(sqrt(2)) - 1 = 2sqrt(2) - 6sqrt(2) - 1 = -4sqrt(2) - 1 For x = -sqrt(2): g(-sqrt(2)) = (-sqrt(2))^3 - 6(-sqrt(2)) - 1 = -2sqrt(2) + 6sqrt(2) - 1 = 4sqrt(2) - 1 Step 4: Identify P and Q from the graph. P is the local maximum and Q is the local minimum. The coordinates of P are (-sqrt(2), 4sqrt(2) - 1) and the coordinates of Q are (sqrt(2), -4sqrt(2) - 1). 9.3 Calculate the values of x for which g strictly increases as x increases. The function g(x) strictly increases when its first derivative g'(x) > 0. Step 1: Set g'(x) > 0. 3x^2 - 6 > 0 3x^2 > 6 x^2 > 2 Step 2: Solve the inequality for x. x < -sqrt(2) or x > sqrt(2) The values of x for which g strictly increases are x (-, -sqrt(2)) (sqrt(2), ). 9.4 If h(x) = g'(x), determine for which values of x is h(x) 0. This means finding the values of x for which g'(x) 0. Step 1: Set g'(x) 0. 3x^2 - 6 0 3x^2 6 x^2 2 Step 2: Solve the inequality for x. -sqrt(2) x sqrt(2) The values of x for which h(x) 0 are x [-sqrt(2), sqrt(2)]. 9.5 Determine the equation of the tangent to g at R. The coordinates of R are (0, -1) from part 9.1. The gradient of the tangent at R is g'(0). Step 1: Calculate the gradient m = g'(0). g'(x) = 3x^2 - 6 m = g'(0) = 3(0)^2 - 6 = -6 Step 2: Use the point-slope form of a linear equation, y - y_1 = m(x - x_1). y - (-1) = -6(x - 0) y + 1 = -6x y = -6x - 1 The equation of the tangent to g at R is y = -6x - 1. 9.6 Write down the equation of the line perpendicular to the tangent at P. The coordinates of P are (-sqrt(2), 4sqrt(2) - 1) from part 9.2. P is a turning point, so the tangent at P is a horizontal line. Step 1: The gradient of the tangent at P is g'(-sqrt(2)) = 0. Step 2: A line perpendicular to a horizontal line (gradient 0) is a vertical line. Step 3: A vertical line passing through P (-sqrt(2), 4sqrt(2) - 1) has the equation x = -sqrt(2). The equation of the line perpendicular to the tangent at P is x = -sqrt(2). Send me the next one 📸