At what value of x is the function f(x) = x(1 | vedclass

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(C) Let $f(x) = x - x^3$. To find the critical points,we find the derivative $f'(x) = 1 - 3x^2$. Setting $f'(x) = 0$,we get $1 - 3x^2 = 0$,which impli

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$x = 1/\sqrt{3}$

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(C) Let $f(x) = x - x^3$. To find the critical points,we find the derivative $f'(x) = 1 - 3x^2$. Setting $f'(x) = 0$,we get $1 - 3x^2 = 0$,which implies $x^2 = 1/3$,so $x = \pm 1/\sqrt{3}$. Since the interval is $0 \leq x \leq 2$,we only consider $x = 1/\sqrt{3}$. Now,we check the second derivative $f''(x) = -6x$. At $x = 1/\sqrt{3}$,$f''(1/\sqrt{3}) = -6/\sqrt{3} Since the second derivative is negative,the function has a local maximum at $x = 1/\sqrt{3}$. We also check the endpoints of the interval $[0, 2]$: $f(0) = 0(1 - 0) = 0$. $f(2) = 2(1 - 4) = 2(-3) = -6$. $f(1/\sqrt{3}) = (1/\sqrt{3})(1 - 1/3) = (1/\sqrt{3})(2/3) = 2/(3\sqrt{3}) \approx 0.385$. Comparing the values $0$,$-6$,and $2/(3\sqrt{3})$,the maximum value occurs at $x = 1/\sqrt{3}$.

At what value of $x$ is the function $f(x) = x(1 - x^2)$ maximum for $0 \leq x \leq 2$?

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