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Chapter 3: Problem 64

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

Short Answer

Expert verified
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 by step solution

01

Finding the Derivatives

Find \(f'(x)\) and \(f''(x)\) by applying the product rule and the power rule of differentiation. \[f'(x) = (2x-6)x + (x-6)^2\]\[f''(x) = 2x + 4(x-6)\]
02

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.
03

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.
04

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 \(f(x)=x(x-6)^{2}\).

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