(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 .

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Let's analyze the function $$ f(x) = x^{3} - 3x^{2} - 24x $$ and address each part of the problem step by step. --- ### **Part (a): Finding the Coordinates of the Local Maximum and Local Minimum** To find the local extrema (maximum and minimum points) of the function $$ f(x) $$, we need to determine its critical points by finding where the first derivative $$ f'(x) $$ is zero or undefined. Then, we'll use the second derivative test to classify these critical points. 1. **First Derivative $$ f'(x) $$:** $$ f'(x) = \frac{d}{dx} \left( x^{3} - 3x^{2} - 24x \right) = 3x^{2} - 6x - 24 $$ 2. **Finding Critical Points:** Set $$ f'(x) = 0 $$ and solve for $$ x $$: $$ 3x^{2} - 6x - 24 = 0 $$ Divide the equation by 3 to simplify: $$ x^{2} - 2x - 8 = 0 $$ Factoring or using the quadratic formula: $$ x = \frac{2 \pm \sqrt{(2)^2 - 4(1)(-8)}}{2(1)} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2} $$ Thus, the critical points are: $$ x = \frac{2 + 6}{2} = 4 \quad \text{and} \quad x = \frac{2 - 6}{2} = -2 $$ 3. **Second Derivative $$ f''(x) $$:** $$ f''(x) = \frac{d}{dx} \left( 3x^{2} - 6x - 24 \right) = 6x - 6 $$ - **At $$ x = 4 $$:** $$ f''(4) = 6(4) - 6 = 24 - 6 = 18 > 0 $$ Since $$ f''(4) > 0 $$, $$ x = 4 $$ corresponds to a **local minimum**. - **At $$ x = -2 $$:** $$ f''(-2) = 6(-2) - 6 = -12 - 6 = -18 < 0 $$ Since $$ f''(-2) < 0 $$, $$ x = -2 $$ corresponds to a **local maximum**. 4. **Coordinates of the Extrema:** - **Local Maximum at $$ x = -2 $$:** $$ f(-2) = (-2)^3 - 3(-2)^2 - 24(-2) = -8 - 12 + 48 = 28 $$ So, the local maximum point is $$ (-2, 28) $$. - **Local Minimum at $$ x = 4 $$:** $$ f(4) = (4)^3 - 3(4)^2 - 24(4) = 64 - 48 - 96 = -80 $$ So, the local minimum point is $$ (4, -80) $$. --- ### **Part (b): Showing the Curve Intersects the $$ x $$-Axis at $$ x = 0 $$** To find the points where the curve intersects the $$ x $$-axis, we set $$ f(x) = 0 $$ and solve for $$ x $$. $$ f(x) = x^{3} - 3x^{2} - 24x = 0 $$ Factor out an $$ x $$: $$ x(x^{2} - 3x - 24) = 0 $$ This gives us: $$ x = 0 \quad \text{or} \quad x^{2} - 3x - 24 = 0 $$ Thus, $$ x = 0 $$ is one of the roots. Therefore, the curve intersects the $$ x $$-axis at $$ x = 0 $$. --- ### **Part (c): Rough Sketch of the Curve** To sketch the curve of $$ f(x) = x^{3} - 3x^{2} - 24x $$, consider the following key features: 1. **Intercepts:** - **$$ x $$-intercepts:** Found by setting $$ f(x) = 0 $$. We have already established that $$ x = 0 $$ is a root. Let's find the other roots from: $$ x^{2} - 3x - 24 = 0 \implies x = \frac{3 \pm \sqrt{9 + 96}}{2} = \frac{3 \pm \sqrt{105}}{2} $$ Approximately: $$ \sqrt{105} \approx 10.247 \implies x \approx \frac{3 \pm 10.247}{2} \implies x \approx 6.623 \quad \text{and} \quad x \approx -3.623 $$ Therefore, the $$ x $$-intercepts are at $$ x \approx -3.62 $$, $$ x = 0 $$, and $$ x \approx 6.62 $$. - **$$ y $$-intercept:** When $$ x = 0 $$, $$ f(0) = 0 $$. So, the point is $$ (0, 0) $$. 2. **Local Extrema:** - **Local Maximum:** $$ (-2, 28) $$ - **Local Minimum:** $$ (4, -80) $$ 3. **Behavior of the Function:** - Since it's a cubic function with a positive leading coefficient, as $$ x \to \infty $$, $$ f(x) \to \infty $$, and as $$ x \to -\infty $$, $$ f(x) \to -\infty $$. - The function has two turning points: a local maximum at $$ (-2, 28) $$ and a local minimum at $$ (4, -80) $$. - The function crosses the $$ x $$-axis at three points: approximately $$ x = -3.62 $$, $$ x = 0 $$, and $$ x = 6.62 $$. This aligns with the information that the curve crosses between $$ -4 $$ and $$ -3 $$, and between $$ 6 $$ and $$ 7 $$. 4. **Sketching the Curve:** Here's a rough sketch based on the above information: ![Graph of f(x) = x³ - 3x² - 24x](https://i.imgur.com/XXXXXXX.png) *(As this is a text-based response, please visualize or plot the function using graphing software with the following key points:* - **Intercepts:** $$ (-3.62, 0) $$, $$ (0, 0) $$, $$ (6.62, 0) $$ - **Local Maximum:** $$ (-2, 28) $$ - **Local Minimum:** $$ (4, -80) $$ - **End Behavior:** Downward to the left and upward to the right. --- ### **Summary of Key Points for the Sketch:** - **Intercepts:** - $$ x $$-intercepts at $$ x \approx -3.62 $$, $$ 0 $$, and $$ 6.62 $$ - $$ y $$-intercept at $$ (0, 0) $$ - **Local Extrema:** - Local Maximum at $$ (-2, 28) $$ - Local Minimum at $$ (4, -80) $$ - **End Behavior:** - $$ f(x) \to -\infty $$ as $$ x \to -\infty $$ - $$ f(x) \to \infty $$ as $$ x \to \infty $$ Use these points and behaviors to draw a smooth cubic curve that accurately reflects the function's properties. **Part (a):** - **Local Maximum:** At $$ x = -2 $$, $$ y = 28 $$ - **Local Minimum:** At $$ x = 4 $$, $$ y = -80 $$ **Part (b):** The curve intersects the $$ x $$-axis at $$ x = 0 $$. **Part (c):** The curve crosses the $$ x $$-axis between $$ -4 $$ and $$ -3 $$, and between $$ 6 $$ and $$ 7 $$.

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