Given: nbsp;fx=x32x 13⇒f'x=3x22x 13+6x32x 12For nbsp;the nbsp;local nbsp;maxima nbsp;or nbsp;minima, nbsp;we nbsp;must nbsp;have nbsp;f'x=0⇒ ...

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Second Derivative Test for Local Minimum

fx=x32 x-13

Question

f(x) = x³ (2x $-$ 1)³

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Solution

$Given : f (x) = x^{3} {(2 x - 1)}^{3} \Rightarrow f' (x) = 3 x^{2} {(2 x - 1)}^{3} + 6 x^{3} {(2 x - 1)}^{2} For the local maxima or minima, we must have f' (x) = 0 \Rightarrow 3 x^{2} {(2 x - 1)}^{3} + 6 x^{3} {(2 x - 1)}^{2} = 0 \Rightarrow 3 x^{2} {(2 x - 1)}^{2} (2 x - 1 + 2 x) = 0 \Rightarrow x^{2} {(2 x - 1)}^{2} (4 x - 1) = 0 \Rightarrow x = 0, \frac{1}{2} and \frac{1}{4}$

Since f '(x) changes from negative to positive when x increases through $\frac{1}{4}$ , x = $\frac{1}{4}$ is a point of local minima.
The local minimum value of f (x) at x = $\frac{1}{4}$ is given by
${(\frac{1}{4})}^{3} {(\frac{1}{2} - 1)}^{3} = \frac{- 1}{512}$

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