The real number which most exceeds its cube i | vedclass

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(B) Let the number be $x$. We want to maximize the difference between the number and its cube,so we define the function $f(x) = x - x^3$.To find the m

Vedclass · Accepted Answer

$1/\sqrt{3}$

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(B) Let the number be $x$. We want to maximize the difference between the number and its cube,so we define the function $f(x) = x - x^3$.To find the maximum,we take the first derivative: $f'(x) = 1 - 3x^2$.Setting $f'(x) = 0$,we get $1 - 3x^2 = 0$,which implies $x^2 = 1/3$,so $x = \pm 1/\sqrt{3}$.Now,we check the second derivative: $f''(x) = -6x$.For $x = 1/\sqrt{3}$,$f''(1/\sqrt{3}) = -6/\sqrt{3} For $x = -1/\sqrt{3}$,$f''(-1/\sqrt{3}) = 6/\sqrt{3} > 0$,which indicates a local minimum.Thus,the number that most exceeds its cube is $1/\sqrt{3}$.

The real number which most exceeds its cube is

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