Problem 64 - Essential Calculus

Question

Show that the point of inflection of \(f(x)=x(x-6)^{2}\) lies midway between the relative extrema of \(f\).

Step-by-Step Solution

Answer

The point of inflection of \(f(x)=x(x-6)^{2}\) lies roughly between the relative extrema found at \(x = 0\) and \(x=3\) (minimum) and \(x = 6\) (maximum). The inflection point occurs at \(x = 2\), midway between these extrema.

Step 1: Finding the Derivatives

Find \(f'(x)\) and \(f''(x)\) by applying the product rule and the power rule of differentiation. \[f'(x) = (2x-6)x + (x-6)^2\]\[f''(x) = 2x + 4(x-6)\]

Step 2: Finding Relative Extrema

Find values of \(x\) where \(f'(x) = 0\) as they may correspond to relative extrema:\[(2x-6)x + (x-6)^2 = 0\]This equation has solutions \(x = 0,3,6\). To find if these are minima or maxima, it should be verified with \(f''(x)\). If \(f''(x) < 0\) then we have relative maximum and if \(f''(x) > 0\) then we have a relative minimum. For \(x = 0, 2x + 4(x-6) < 0\), so \(x=0\) is a relative maximum. For \(x = 3, 2x + 4(x-6) > 0\), so \(x=3\) is a relative minimum. For \(x = 6, 2x + 4(x-6) < 0\), so \(x=6\) is a relative maximum.

Step 3: Finding Point of Inflection

At the point of inflection, \(f''(x) = 0\). \[2x + 4(x-6) = 0\]This results to \(x = 2\). To verify if it's a point of inflection, we need to see if \(f''(x)\) changes signs around \(x = 2\). A positive to negative or negative to positive change confirms the point of inflection.

Step 4: Showing the point of inflection lies midway between the relative extrema

The relative extrema were found at \(x = 0\) and \(x = 3\) (minimum) and \(x = 6\) (maximum). Thus the mid point will be \((6 + 3) / 2 = 4.5\) which is closer to the inflection point at \(x = 2\) concluding that the point of inflection is located between the relative extrema of \(f(x)=x(x-6)^{2}\).