Word Count : 625

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To find the extreme values of the function \( f(x) = (x+1)^3 (x-3)^2 \), we need to follow these steps: 1. Find the first derivative \( f'(x) \). 2. Set \( f'(x) = 0 \) and solve for \( x \) to find critical points. 3. Determine the nature of the critical points (whether they are maxima, minima, or points of inflection). --- ### Step 1: Find the first derivative \( f'(x) \) The function is \( f(x) = (x+1)^3 (x-3)^2 \). To find \( f'(x) \), we use the product rule: \[ f'(x) = \frac{d}{dx} \left[ (x+1)^3 \right] \cdot (x-3)^2 + (x+1)^3 \cdot \frac{d}{dx} \left[ (x-3)^2 \right]. \] Now, compute the derivatives of each part: \[ \frac{d}{dx} \left[ (x+1)^3 \right] = 3(x+1)^2, \] \[ \frac{d}{dx} \left[ (x-3)^2 \right] = 2(x-3). \] Substitute these into the product rule: \[ f'(x) = 3(x+1)^2 (x-3)^2 + (x+1)^3 \cdot 2(x-3). \] Factor out the common terms: \[ f'(x) = _________ __________ __________ _________ ________ ____ _________ __________ ______ _______.
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