## Step 1: Find the Relative Extrema First, let's find the **relative extrema** by setting the first derivative to zero. Given: \[ y = x(x-6)^2 \] Find the first derivative: \[ y' = \frac{d}{dx}[x(x-6)^2] \] Apply the product rule: \[ y' = (x)'(x-6)^2 + x \cdot 2(x-6) \] \[ y' = (x-6)^2 + 2x(x-6) \] \[ y' = (x-6)^2 + 2x(x-6) \] Factor \( (x-6) \): \[ y' = (x-6)[(x-6) + 2x] = (x-6)(x-6 + 2x) = (x-6)(3x - 6) \] 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 Inflection Point Now, let's find the **point of inflection** by setting the second derivative to zero. Recall: \[ y' = (x-6)(3x-6) \] Expand: \[ y' = 3x^2 - 18x + 36 \] Find the second derivative: \[ y'' = \frac{d}{dx}(3x^2 - 18x + 36) = 6x - 18 \] Set \( y'' = \): \[ 6x - 18 = \implies x = 3 \] **Explanation:** The point of inflection is at \( x = 3 \). --- ## Final Answer The point of inflection (\( x = 3 \)) is exactly **midway** between the relative extrema (\( x = 2 \) and \( x = 6 \)), since the midpoint of 2 and 6 is \( \frac{2+6}{2} = 4 \). However, the calculation shows the inflection point is at \( x = 3 \). Let's double-check the midpoint: - Distance between extrema: \( 6 - 2 = 4 \) - Midpoint: \( 2 + \frac{4}{2} = 4 \) But the inflection point is at \( x = 3 \), not 4. Upon reviewing the calculations, let's verify the first derivative step: Original function: \( y = x(x-6)^2 \) First derivative: \[ y' = (x-6)^2 + x \cdot 2(x-6) = (x-6)^2 + 2x(x-6) = (x-6)[(x-6) + 2x] = (x-6)(x-6+2x) = (x-6)(3x-6) \] Second derivative: \[ y' = (x-6)(3x-6) \] Use product rule: \[ y'' = (x-6)'(3x-6) + (x-6)(3x-6)' = (1)(3x-6) + (x-6)(3) = 3x-6 + 3(x-6) = 3x-6 + 3x - 18 = 6x - 24 \] Set \( y'' = \): \[ 6x - 24 = \implies x = 4 \] **Correction:** The point of inflection is at \( x = 4 \), which **is** midway between \( x = 2 \) and \( x = 6 \). --- ### **Final Answer** **The point of inflection (\( x = 4 \)) is exactly midway between the relative extrema (\( x = 2 \) and \( x = 6 \)).**

# Solution to the Problem ## Step 1: Find the Relative Extrema To find the **relative extrema**, we start with the function: \[ y = x(x-6)^2 \] ### Finding the First Derivative Using the product rule, we calculate the first derivative: \[ y' = \frac{d}{dx}[x(x-6)^2] \] Applying the product rule: \[ y' = (x)'(x-6)^2 + x \cdot 2(x-6) \] This simplifies to: \[ y' = (x-6)^2 + 2x(x-6) = (x-6)[(x-6) + 2x] = (x-6)(3x - 6) \] ### Setting the First Derivative to Zero Now, set \( y' = 0 \): \[ (x-6)(3x-6) = 0 \] This gives us two solutions: 1. \( x - 6 = 0 \) → \( x = 6 \) 2. \( 3x - 6 = 0 \) → \( x = 2 \) **Explanation:** The relative extrema occur at \( x = 2 \) and \( x = 6 \). --- ## Step 2: Find the Inflection Point Next, we find the **point of inflection** by calculating the second derivative. ### Finding the Second Derivative Recall: \[ y' = (x-6)(3x-6) \] Expanding the first derivative: \[ y' = 3x^2 - 18x + 36 \] Now, we differentiate again: \[ y'' = \frac{d}{dx}(3x^2 - 18x + 36) = 6x - 18 \] ### Setting the Second Derivative to Zero Set \( y'' = 0 \): \[ 6x - 18 = 0 \implies x = 3 \] **Explanation:** The point of inflection is at \( x = 3 \). --- ## Final Answer The point of inflection (\( x = 3 \)) does not lie midway between the relative extrema (\( x = 2 \) and \( x = 6 \)). The midpoint of these extrema is: \[ \frac{2 + 6}{2} = 4 \] However, the calculations show that the inflection point is at \( x = 3 \). ### **Conclusion** - **Relative Extrema**: \( x = 2 \) and \( x = 6 \) - **Point of Inflection**: \( x = 3 \) (not midway between the extrema). Thus, the final answer remains: **The point of inflection is at \( x = 3 \)**, which is not midway between the relative extrema \( x = 2 \) and \( x = 6 \).