# Step 1: Find the Relative Extrema Given function: \[ y = x(x-6)^2 \] First, find the critical points (where \( y' = \)): 1. Compute the first derivative: \[ y = x(x-6)^2 \] Use the product rule: \[ y' = (x)'\cdot(x-6)^2 + x\cdot[2(x-6)] \] \[ = (x-6)^2 + 2x(x-6) \] \[ = (x-6)^2 + 2x(x-6) \] \[ = (x-6)[(x6) + 2x] \] \[ = (x-6)(x-6+2x) \] \[ = (x-6)(3x-6) \] \[ = 3x^2 - 6x - 18x + 36 \] \[ = 3x^2 - 24x + 36 \] Or just keep as \( (x-6)(3x-6) \). 2. Set \( y' = \): \[ (x-6)(3x-6) = \] So, \[ x-6 = \implies x = 6 \] \[ 3x-6 = \implies x = 2 \] **Explanation:** The relative extrema occur at \( x = 2 \) and \( x = 6 \). --- # Step 2: Find the Point of Inflection 1. Compute the second derivative: Recall: \[ y' = (x-6)(3x-6) \] Use the product rule again: \[ y'' = (x-6)'(3x-6) + (x-6)(3x-6)' \] \[ = (1)(3x-6) + (x-6)(3) \] \[ = 3x - 6 + 3(x-6) \] \[ = 3x - 6 + 3x - 18 \] \[ = 6x - 24 \] 2. Set \( y'' = \): \[ 6x - 24 = \] \[ x = 4 \] **Explanation:** The point of inflection is at \( x = 4 \). --- # Final Answer The point of inflection (\( x = 4 \)) is exactly midway between the relative extrema (\( x = 2 \) and \( x = 6 \)), since \( 4 \) is the average of \( 2 \) and \( 6 \): \[ \frac{2+6}{2} = 4 \]

# Step 1: Find the Relative Extrema Given function: \[ y = x(x-6)^2 \] First, find the critical points by calculating the derivative and setting it to zero: 1. **Compute the first derivative** using the product rule: \[ y' = (x)'\cdot(x-6)^2 + x\cdot[2(x-6)(x)'] \] \[ = (x-6)^2 + 2x(x-6) \] \[ = (x-6)[(x-6) + 2x] \] \[ = (x-6)(3x-6) \] 2. **Set \( y' = 0 \)**: \[ (x-6)(3x-6) = 0 \] This gives us two equations: \[ x-6 = 0 \implies x = 6 \] \[ 3x-6 = 0 \implies x = 2 \] **Explanation:** The relative extrema occur at \( x = 2 \) and \( x = 6 \). --- # Step 2: Find the Point of Inflection 1. **Compute the second derivative**: Recall: \[ y' = (x-6)(3x-6) \] Using the product rule again: \[ y'' = (x-6)'(3x-6) + (x-6)(3x-6)' \] \[ = (1)(3x-6) + (x-6)(3) \] \[ = 3x - 6 + 3(x-6) \] \[ = 3x - 6 + 3x - 18 \] \[ = 6x - 24 \] 2. **Set \( y'' = 0 \)**: \[ 6x - 24 = 0 \] \[ x = 4 \] **Explanation:** The point of inflection is at \( x = 4 \). --- # Final Answer The relative extrema occur at \( x = 2 \) and \( x = 6 \). The point of inflection is at \( x = 4 \). This point is exactly midway between the relative extrema.