(11) \( f(x)=x^{3}-3 x^{2}-24 x, x \in \mathbb{R} \) (a) Find the co-ordinates of the local maximum point and the local minimum point on the curve of \( y=f(x) \). (b) Show that this curve intersects the \( x \)-axis at \( x=0 \). (c) Draw a rough sketch of the curve given that it crosses the \( x \)-axis between -4 and -3 and also between 6 and 7 .

To find the local maximum and minimum points of the function \( f(x) = x^{3} - 3x^{2} - 24x \), we start by finding its derivative, \( f'(x) \). We have: \[ f'(x) = 3x^{2} - 6x - 24. \] Setting the derivative to zero to find critical points gives: \[ 3x^{2} - 6x - 24 = 0 \] \[ x^{2} - 2x - 8 = 0 \] \[ (x-4)(x+2) = 0. \] Thus, \( x = 4 \) and \( x = -2 \) are critical points. To determine whether these points are local maxima or minima, we evaluate the second derivative: \[ f''(x) = 6x - 6. \] For \( x = 4 \): \[ f''(4) = 6(4) - 6 = 24 > 0 \quad \text{(local minimum)}. \] For \( x = -2 \): \[ f''(-2) = 6(-2) - 6 = -18 < 0 \quad \text{(local maximum)}. \] Now, let's calculate the function values at these critical points: - For \( x = 4 \): \[ f(4) = 4^{3} - 3(4^{2}) - 24(4) = 64 - 48 - 96 = -80. \] - For \( x = -2 \): \[ f(-2) = (-2)^{3} - 3(-2)^{2} - 24(-2) = -8 - 12 + 48 = 28. \] So the local maximum point is at \( (-2, 28) \) and the local minimum point is at \( (4, -80) \). For part (b), we show that the curve intersects the \( x \)-axis at \( x = 0 \): \[ f(0) = 0^{3} - 3(0)^{2} - 24(0) = 0. \] Thus, the curve indeed intersects the \( x \)-axis at the point \( (0,0) \). For part (c), you can expect the curve to look like a typical cubic function with a local maximum at \( (-2, 28) \) and local minimum at \( (4, -80) \). Given that it crosses the \( x \)-axis between -4 and -3 as well as between 6 and 7, sketch it accordingly, showing a drop from the local maximum going down to intersect the axis, rising again to the local minimum, and then falling off again to eventually cross the axis again in the specified ranges.