Let's solve **Question 1.1** step-by-step. --- ### 1.1. \( f(x) = 3 \sqrt[3]{x} - 16x^2 \) for \( -1 \leq x \leq 1 \) --- #### **a) Find the derivative, specify its domain, and determine all critical points.** **Step 1: Find the derivative** - \( f(x) = 3x^{1/3} - 16x^2 \) - Derivative: \[ f'(x) = 3 \cdot \frac{1}{3} x^{-2/3} - 32x \] \[ f'(x) = x^{-2/3} - 32x \] \[ f'(x) = \frac{1}{x^{2/3}} - 32x \] **Step 2: Domain of the derivative** - \( x^{2/3} \) is defined for all real \( x \) (cube root is defined for negatives, square of cube root is always non-negative). - But, \( x^{-2/3} \) is undefined at \( x = 0 \). - So, domain of \( f(x) \) is \( [-1, 1] \), but domain of \( f'(x) \) is \( [-1, 0) \cup (0, 1] \). **Step 3: Critical points** Critical points occur where \( f'(x) = 0 \) or is **undefined** (within the domain of \( f(x) \)). - \( f'(x) \) is **undefined at \( x = 0 \)** (singular point). - \( f'(x) = 0 \): \[ \frac{1}{x^{2/3}} - 32x = 0 \] \[ \frac{1}{x^{2/3}} = 32x \] \[ 1 = 32x^{5/3} \] \[ x^{5/3} = \frac{1}{32} \] \[ x = \left(\frac{1}{32}\right)^{3/5} \] \[ x = \left(2^{-5}\right)^{3/5} = 2^{-3} = \frac{1}{8} \] - So, critical points: \( x = 0 \) (singular), \( x = \frac{1}{8} \) (in interval). --- #### **b) Intervals of increase and decrease** Check the sign of \( f'(x) \): - **For \( x < 0 \):** \( x^{-2/3} \) positive, \( -32x \) positive (since \( x \) negative, so minus times minus is plus). So \( f'(x) = \text{positive} + \text{positive} > 0 \), function is **increasing** on \( [-1, 0) \). - **For \( 0 < x < \frac{1}{8} \):** - \( x^{-2/3} \) positive, \( -32x \) negative, but \( x^{-2/3} \) dominates for small \( x \). - Try \( x = 0.01 \): \( 1/(0.01^{2/3}) \) is very large, minus a small value. So \( f'(x) > 0 \) near 0. - **At \( x = \frac{1}{8} \):** \( f'(x) = 0 \) - **For \( x > \frac{1}{8} \):** \( x^{-2/3} \) decreases, \( -32x \) gets more negative as \( x \) increases. Eventually, \( -32x \) dominates, so \( f'(x) < 0 \). - Try \( x = 0.5 \): \( 1/(0.5^{2/3}) \approx 1.59 \), \( -32(0.5) = -16 \), so \( f'(x) \approx 1.59 - 16 < 0 \). **Summary:** - **Increasing on:** \( [-1, 0) \cup (0, \frac{1}{8}) \) - **Decreasing on:** \( (\frac{1}{8}, 1] \) --- #### **c) Stationary, singular, and endpoints (local/absolute min/max)** - **Stationary point:** \( x = \frac{1}{8} \) - **Singular point:** \( x = 0 \) - **Endpoints:** \( x = -1, x = 1 \) Evaluate \( f(x) \) at these: - \( f(-1) = 3(-1) - 16(1) = -3 - 16 = -19 \) - \( f(0) = 0 - 0 = 0 \) - \( f(1/8) = 3(1/8)^{1/3} - 16(1/8)^2 \) - \( (1/8)^{1/3} = 1/2 \) - So, \( f(1/8) = 3 \times 1/2 - 16 \times (1/64) = 1.5 - 0.25 = 1.25 \) - \( f(1) = 3(1) - 16(1) = 3 - 16 = -13 \) **Maximum value:** \( f(1/8) = 1.25 \) (**local and absolute maximum**) **Minimum value:** \( f(-1) = -19 \) (**absolute minimum**) --- ### **Final Answers (for 1.1):** --- #### a) **Derivative and critical points:** - \( f'(x) = \frac{1}{x^{2/3}} - 32x \) - Domain of \( f'(x) \): \( [-1, 0) \cup (0, 1] \) - Critical points: \( x = 0 \) (singular), \( x = \frac{1}{8} \) (stationary) #### b) **Intervals of increase/decrease:** - Increasing on \( [-1, 0) \cup (0, \frac{1}{8}) \) - Decreasing on \( (\frac{1}{8}, 1] \) #### c) **Stationary, singular, endpoints and extrema:** - **Stationary:** \( x = \frac{1}{8} \), value \( f(1/8) = 1.25 \) (**local/absolute max**) - **Singular:** \( x = 0 \), value \( f(0) = 0 \) - **Endpoints:** \( x = -1 \), value \( f(-1) = -19 \) (**absolute min**); \( x = 1 \), value \( f(1) = -13 \) --- Would you like a solution for **1.2** as well?

Let's solve **Question 1.2** step-by-step. --- ### 1.2. \( g(x) = (\sin x)^3 \) for \( -\pi < x < \pi \) --- #### **a) Find the derivative, specify its domain, and determine all critical points.** **Step 1: Find the derivative** Given: \[ g(x) = (\sin x)^3 \] Using the chain rule: \[ g'(x) = 3 (\sin x)^2 \cdot \cos x \] **Derivative:** \[ g'(x) = 3 \sin^2 x \cdot \cos x \] **Domain of \( g(x) \):** \( (-\pi, \pi) \), which is open interval, and \( g'(x) \) is defined everywhere except at points where \( \sin x \) or \( \cos x \) might cause issues, but since both are continuous and defined on \( (-\pi, \pi) \), the derivative exists everywhere in the domain. --- #### **b) Write the intervals where the function is increasing or decreasing.** **Step 2: Find critical points where \( g'(x) = 0 \) or undefined** - \( g'(x) = 0 \) when: \[ 3 \sin^2 x \cdot \cos x = 0 \] \[ \Rightarrow \sin^2 x = 0 \quad \text{or} \quad \cos x = 0 \] - \( \sin^2 x = 0 \) when \( \sin x = 0 \): \[ x = 0, \pm \pi \] But since the domain is \( (-\pi, \pi) \), the endpoints are not included, so critical point at \( x=0 \). - \( \cos x = 0 \): \[ x = \pm \frac{\pi}{2} \] Both are in \( (-\pi, \pi) \). **Critical points:** \[ x = 0, \pm \frac{\pi}{2} \] --- **Step 3: Determine the sign of \( g'(x) \) in the intervals** Intervals: - \( (-\pi, -\frac{\pi}{2}) \) - \( (-\frac{\pi}{2}, 0) \) - \( (0, \frac{\pi}{2}) \) - \( (\frac{\pi}{2}, \pi) \) Choose test points in each: - \( x = -\frac{3\pi}{4} \): - \( \sin x < 0 \), \( \sin^2 x > 0 \), \( \cos x < 0 \) - \( g'(x) = 3 \times (\text{positive}) \times (\text{negative}) < 0 \) - **Decreasing** - \( x = -\frac{\pi}{4} \): - \( \sin x < 0 \), \( \sin^2 x > 0 \), \( \cos x > 0 \) - \( g'(x) > 0 \) - **Increasing** - \( x = \frac{\pi}{4} \): - \( \sin x > 0 \), \( \sin^2 x > 0 \), \( \cos x > 0 \) - \( g'(x) > 0 \) - **Increasing** - \( x = \frac{3\pi}{4} \): - \( \sin x > 0 \), \( \sin^2 x > 0 \), \( \cos x < 0 \) - \( g'(x) < 0 \) - **Decreasing** --- ### **Summary of increase/decrease:** - **Increasing:** \( (-\frac{\pi}{2}, 0) \) and \( (0, \frac{\pi}{2}) \) - **Decreasing:** \( (-\pi, -\frac{\pi}{2}) \) and \( (\frac{\pi}{2}, \pi) \) --- #### **c) Find the stationary points, singular points, endpoints, and classify the local/absolute extrema.** - **Stationary points:** where \( g'(x) = 0 \): \[ x = 0, \pm \frac{\pi}{2} \] - **Endpoints:** The interval is open, so endpoints \( -\pi \) and \( \pi \) are not included. **Evaluate \( g(x) \) at critical points:** - \( g(0) = (\sin 0)^3 = 0 \) - \( g(\pm \frac{\pi}{2}) = (\sin \pm \frac{\pi}{2})^3 = ( \pm 1)^3 = \pm 1 \) --- ### **Determine maxima and minima:** - **At \( x = -\frac{\pi}{2} \):** \( g(-\frac{\pi}{2}) = -1 \) - **At \( x = 0 \):** \( g(0) = 0 \) - **At \( x = \frac{\pi}{2} \):** \( g(\frac{\pi}{2}) = 1 \) --- ### **Summary:** | Point | \( g(x) \) | Type | Explanation | |---------|--------------|---------|----------------| | \( x = -\frac{\pi}{2} \) | \( -1 \) | Local **minimum** | Function decreases before and increases after | | \( x = 0 \) | 0 | Neither max nor min | Change in increasing/decreasing pattern | | \( x = \frac{\pi}{2} \) | 1 | Local **maximum** | Function increases before and decreases after | --- ### **Final results:** - **Critical points:** \( x = 0, \pm \frac{\pi}{2} \) - **Intervals of increase:** \( (-\frac{\pi}{2}, 0) \) and \( (0, \frac{\pi}{2}) \) - **Intervals of decrease:** \( (-\pi, -\frac{\pi}{2}) \) and \( (\frac{\pi}{2}, \pi) \) - **Local maxima:** at \( x = \frac{\pi}{2} \), \( g = 1 \) - **Local minima:** at \( x = -\frac{\pi}{2} \), \( g = -1 \) --- **Would you like a summary or any further clarification?**