Find-area-of-square-inserted-in-curve-f-x-3x-x-3-

Question Number 41461 by rahul 19 last updated on 07/Aug/18

Find area of square inserted in curve

f (x) = 3 x - x^{3} .

Commented by rahul 19 last updated on 07/Aug/18

Commented by rahul 19 last updated on 07/Aug/18

Ans : {(216)}^{\frac{1}{3}} + {(- 108)}^{\frac{1}{3}} .

Commented by MJS last updated on 07/Aug/18

216^{\frac{1}{3}} = 6

{(- 108)}^{\frac{1}{3}} = - 108^{\frac{1}{3}} = - {(2^{2} 3^{3})}^{\frac{1}{3}} = - 3 \sqrt[3]{4}

\Rightarrow answer = 6 - 3 \sqrt[3]{4}

Answered by MJS last updated on 07/Aug/18

idea : intersect y = d with f (x)

we {don}^{'} t need x_{1} < 0

solve x_{3} - x_{2} = d for d

3 x - x^{3} = d

x^{3} - 3 x + d = 0

[f^{'} (x) = 3 - 3 x^{2} \Rightarrow \max (f (x)) = (\begin{matrix} 1 \\ 2 \end{matrix}) \Rightarrow

\Rightarrow 0 < d < 2 \Leftrightarrow x_{1}, x_{2}, x_{3} \in R \Rightarrow

\Rightarrow we need the trigonometric method]

x_{1} = - 2 \sin (\frac{1}{3} (π + \arcsin \frac{d}{2}))

x_{2} = 2 \sin (\frac{1}{3} \arcsin \frac{d}{2})

x_{3} = 2 \cos (\frac{1}{3} (\frac{π}{2} + \arcsin \frac{d}{2}))

x_{3} - x_{2} = d

2 \sqrt{3} \cos (\frac{1}{3} (π + \arcsin \frac{d}{2})) = d

\frac{1}{3} (π + \arcsin \frac{d}{2}) = \arccos \frac{d \sqrt{3}}{6}

π + \arcsin \frac{d}{2} = 3 \arccos \frac{d \sqrt{3}}{6}

\tan (π + \arcsin \frac{d}{2}) = \tan (3 \arccos \frac{d \sqrt{3}}{6})

\frac{d}{\sqrt{4 - d^{2}}} = \frac{(d^{2} - 3) \sqrt{12 - d^{2}}}{d (d^{2} - 9)}

this leads to

d^{6} - 18 d^{4} + 108 d^{2} - 108 = 0

d = \sqrt{u}

u^{3} - 18 u^{2} + 108 u - 108 = 0

u = v + 6

v^{3} + 108 = 0

v = - 3 \sqrt[3]{4}

u = 6 - 3 \sqrt[3]{4}

d = \sqrt{6 - 3 \sqrt[3]{4}} \approx 1.11256

area of square = d^{2} = 6 - 3 \sqrt[3]{4} \approx 1.23780

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